Type – I: To Find New Equation of Locus After Shifting the Origin

Example – 01:

The origin is shifted to the point (1, -4). Find the new equation of locus 2x² – xy + 3y² – 8x + 25 y + 54 = 0 axes remaining parallel.

Solution:

The old equation of locus is 2x² – xy + 3y² – 8x + 25 y + 54 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1, -4)

We have x = X + h = X + 1 and y = Y + k = Y – 4

Substituting these values in equation (1) we have

2(X + 1)² – (X + 1)( Y – 4) + 3( Y – 4)² – 8(X + 1) + 25 ( Y – 4) + 54 = 0

∴  2(X² + 2X + 1) – (XY – 4X + Y – 4) + 3( Y² – 8Y + 16) – 8X -8 + 25 Y – 100 + 54 = 0

∴  2X² + 4X + 2 – XY + 4X – Y + 4 + 3Y² – 24Y + 48 – 8X -8 + 25 Y – 100 + 54 = 0

∴  2X² – XY + 3Y² = 0

Ans: The new equation of locus is 2X² – XY + 3Y² = 0

Example – 02:

The origin is shifted to the point (1, 1). Find the new equation of locus xy – x – y + 1 = 0 axes remaining parallel.

Solution:

The old equation of locus is xy – x – y + 1 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1, 1)

We have x = X + h = X + 1 and y = Y + k = Y + 1

Substituting these values in equation (1) we have

(X + 1)( Y + 1) – (X + 1) – (Y + 1) + 1 = 0

∴  XY + X + Y + 1 – X – 1 – Y – 1 + 1 = 0

∴  XY  = 0

Ans: The new equation of locus is XY  = 0

Example – 03:

The origin is shifted to the point (1, 1). Find the new equation of locus x² – y² – 2x + 2y = 0 axes remaining parallel.

Solution:

The old equation of locus is x² – y² – 2x + 2y = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1, 1)

We have x = X + h = X + 1 and y = Y + k = Y + 1

Substituting these values in equation (1) we have

(X + 1)² – (Y + 1)² – 2(X + 1) + 2(Y + 1) = 0

∴  (X² + 2X + 1) – ( Y² + 2Y + 1) – 2X -2 + 2Y  + 2 = 0

∴  X² + 2X + 1 – Y² – 2Y  – 1 – 2X -2 + 2Y  + 2 = 0

∴  X² – Y² = 0

Ans: The new equation of locus is X² – Y² = 0

Example – 04:

The origin is shifted to the point (1, 1). Find the new equation of locus x² + y² – 4x + 6y +3 = 0 axes remaining parallel.

Solution:

The old equation of locus is x² + y² – 4x + 6y +3 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1, 1)

We have x = X + h = X + 1 and y = Y + k = Y + 1

Substituting these values in equation (1) we have

(X + 1)² + (Y + 1)² – 4(X + 1) + 6(Y + 1) + 3 = 0

∴  (X² + 2X + 1) + (Y² + 2Y + 1) – 4X – 4 + 6Y + 6 + 3 = 0

∴  X² + 2X + 1 + Y² + 2Y + 1 – 4X – 4 + 6Y + 6 + 3 = 0

∴  X² + Y² – 2X + 8Y + 7 = 0

Ans: The new equation of locus is X² + Y² – 2X + 8Y + 7 = 0

Example – 05:

The origin is shifted to the point (2, -1). Find the new equation of locus 2x² + 3xy – 9y² – 5x – 24y – 7 = 0 axes remaining parallel.

Solution:

The old equation of locus is 2x² + 3xy – 9y² – 5x – 24y – 7 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (2, -1)

We have x = X + h = X + 2 and y = Y + k = Y – 1

Substituting these values in equation (1) we have

2(X + 2)² + 3(X + 2) (Y – 1) – 9(Y – 1)² – 5(X + 2) – 24(Y – 1) – 7 = 0

∴  2(X² + 4X + 4) + 3(XY – X + 2Y – 2) – 9(Y² – 2Y + 1) – 5X – 10 – 24Y + 24 – 7 = 0

∴  2X² + 8X + 8 + 3XY – 3X + 6Y – 6 – 9Y² + 18Y  – 9 – 5X – 10 – 24Y + 24 – 7 = 0

∴  2X² + 3XY – 9Y² = 0

Ans: The new equation of locus is 2X² + 3XY – 9Y² = 0

Example – 06:

The origin is shifted to the point (a – c, b). Find the new equation of locus (x – a)² + (y – b)² = r² axes remaining parallel.

Solution:

The old equation of locus is (x – a)² + (y – b)² = r² …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (a – c, b)

We have x = X + h = X + a – c and y = Y + k = Y + b

Substituting these values in equation (1) we have

(X + a – c – a)² + (Y + b – b)² = r²

∴  (X – c)² + (Y)² = r²

∴  (X – c)² + (Y)² = r²

∴  X² – 2cX + c² + Y² = r²

∴  X² + Y² – 2cX + c² – r² = 0

Ans: The new equation of locus is X² + Y² – 2cX + c² – r² = 0

Example – 07:

The origin is shifted to the point (1/2, -3/2). Find the new equation of locus (2x – 1)² + (y + 3/2)² = 4 axes remaining parallel.

Solution:

The old equation of locus is (2x – 1)² + (y + 3/2)² = 4  …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (1/2, -3/2)

We have x = X + h = X + 1/2 and y = Y + k = Y – 3/2

Substituting these values in equation (1) we have

(2(X + 1/2) – 1)² + ((Y – 3/2) + 3/2)² = 4

∴  (2X + 1 – 1)² + (Y – 3/2 + 3/2)² = 4

∴  (2X)² + (Y)² = 4

∴  4X² + Y² = 4

Ans: The new equation of locus is 4X² + Y² = 4

Example – 08:

The origin is shifted to the point (-1, 2). Find the new equation of locus 4x² + y² + 6x – 4y + 5 = 0 axes remaining parallel.

Solution:

The old equation of locus is 4x² + y² + 8x – 4y + 4 = 0  …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (2, -1)

We have x = X + h = X – 1 and y = Y + k = Y + 2

Substituting these values in equation (1) we have

4(X – 1)² + (Y + 2)² + 8(X – 1) – 4(Y + 2) + 4 = 0

∴  4(X² – 2X +  1) + (Y² + 4Y + 4) + 8X – 8 – 4Y – 8 + 4 = 0

∴  4X² – 8X +  4 + Y² + 4Y + 4 + 8X – 8 – 4Y – 8 + 4 = 0

∴  4X² + Y² – 4 = 0

∴  4X² + Y² = 4

Ans: The new equation of locus is 4X² + Y² = 4

Example – 09:

Obtain the new equation of locus (a – b)(x² + y²) – 2abx = 0. If the origin is shifted to the point (ab/(a-b), 0) axes remaining parallel.

Solution:

The old equation of locus is (a – b)(x² + y²) – 2abx = 0   …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (2, -1)

We have x = X + h = X + ab/(a-b) and y = Y + k = Y + 0 = Y

Substituting these values in equation (1) we have

Ans: The new equation of locus is (a – b)²(X² + Y²) – a²b² = 0

Type – II: To Find Old Equation of Locus After Shifting the Origin

Example – 10:

The origin is shifted to the point (5, -3). The equation of locus in new system is Y² = 6X. Find the equation of locus in the old system

Solution:

The new equation of locus is Y² = 6X  …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (5, -3)

We have X = x – h = x – 5 and Y = y – k = y + 3

Substituting these values in equation (1) we have

(y + 3)² = 6(x – 5)

∴  y² + 6y + 9 = 6x – 30

∴  y² – 6x + 6y + 39 = 0

Ans: The equation of locus in old system is y² – 6x + 6y + 39 = 0

Example – 11:

The origin is shifted to the point (-1, 2). The equation of locus in new system is X² + 5XY + 3Y² = 0. Find the equation of locus in the old system

Solution:

The new equation of locus is X² + 5XY + 3Y² = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (-1, 2)

We have X = x – h = x + 1 and Y = y – k = y – 2

Substituting these values in equation (1) we have

(x + 1)² + 5(x + 1)(y -2) + 3(y -2)² = 0

∴  (x² + 2x + 1) + 5(xy – 2x + y – 2) + 3(y² – 4y + 4) = 0

∴  x² + 2x + 1 + 5xy – 10x + 5y – 10 + 3y² – 12y + 12 = 0

∴  x² + 5xy + 3y² – 8x – 7y + 3 = 0

Ans: The equation of locus in old system is x² + 5xy + 3y² – 8x – 7y + 3 = 0

Science > Mathematics > Coordinate Geometry > Locus > New Equation of Locus After Shifting Origin

The post New Equation of Locus After Shifting Origin appeared first on The Fact Factor.

Example – 01:

If the equation x² – 4x – 6y + 10 = 0 is transformed to X² + AY = 0 after shifting the origin and  axes remaining parallel. Find the coordinates of the point where the origin is shifted and the value of A.

Solution:

The old equation of locus is x² – 4x – 6y + 10 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

(X + h)² – 4(X + h) – 6(Y + k) + 10 = 0

∴   X² + 2hX + h² – 4X – 4h – 6Y – 6k + 10 = 0

∴   X² + (2h – 4)X – 6Y  + h² – 4h – 6k + 10 = 0 …………. (2)

The new equation of locus is X² + AY = 0

Thus term of X and constant term are absent

∴   2h – 4 = 0 i.e. h = 2

Similarly

h² – 4h – 6k + 10 = 0

∴   (2)² – 4(2) – 6k + 10 = 0

∴   4 – 8 – 6k + 10 = 0

∴    – 6k = – 6

∴    K = 1

Thus the origin is shifted to (2, 1)

Thus the equation (2) reduces to

X² – 6Y = 0

Comparing with given new equation of locus, A = – 6

Ans: origin is shifted to (2,1) and A = -6

Example – 02:

If the equation xy – 3x + 2y – 7 = 0 is transformed to XY = 1 after shifting the origin and axes remaining parallel. Find the coordinates of the point where the origin is shifted.

Solution:

The old equation of locus is xy – 3x + 2y – 7 = 0…………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

(X + h) (Y + k) – 3(X + h) + 2(Y + k) – 7 = 0

∴   XY + kX + hY + hk – 3X – 3h + 2Y + 2k – 7 = 0

∴   XY + (k – 3)X + (h + 2)Y + (hk– 3h + 2k – 7) = 0 …………. (2)

The new equation of locus is XY = 1

Thus terms of X and Y are absent

∴   K – 3 = 0 and h + 2 = 0 and (hk– 3h + 2k – 7) = -1

∴   K = 3 and h = -2

The value of (hk– 3h + 2k – 7) = (-2)(3) – 3(-2) + 2(3) – 7 = -6 + 6 + 6 – 7 = -1

Thus values of h and k satisfy the third relation.

Ans: The origin is shifted to (-2, 3)

Example – 03:

By shifting the origin to a suitable point O’(h, k) axes remaining parallel. The equation 2x² + 2y² -2x + 2y – 1 = 0 reduces to form X² + Y² = a² (a > 0). Find O’(h, k) and value of a.

Solution:

The old equation of locus is 2x² + 2y² -2x + 2y – 1 = 0…………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

2(X + h)² + 2(Y + k)² -2(X + h) + 2(Y + k) – 1 = 0

∴   2(X²+ 2hX + h²) + 2(Y² + 2kY + k²) – 2X – 2h + 2Y + 2k – 1 = 0

∴   2X²+ 4hX + 2h² + 2Y² + 4kY + 2k² – 2X – 2h + 2Y + 2k – 1 = 0

∴   2X² + 2Y² + (4h – 2)X + (4k + 2)Y + 2h² + 2k² – 2h + 2k – 1 = 0

∴   X² + Y² + (2h – 1)X + (2k + 1)Y + h² + k² – h + k – 1/2 = 0  …………. (2)

The new equation of locus is X² + Y² = a²Thus terms of X and Y are absent

∴   2h – 1 = 0 and 2k + 1 = 0 and + h² + k² – h + k – 1/2 = – a²

∴   h = 1/2 and h = -1/2The origin is shifted to (1/2, -1/2)

The value of h² + k² – h + k – 1/2 = – a²

∴   (1/2)² + (-1/2)² – (1/2) + (-1/2) – 1/2 = – a²

∴   1/4 + 1/4 – 1/2 – 1/2 – 1/2 = – a²

∴   – 1/2 – 1/2 = – a²

∴   – 1 = – a²

∴   a² = 1

∴   a = + – 1

∴   a = 1 (since a> 0 given)

Ans: The origin is shifted to O’(1/2, -1/2) and a = 1

Example – 04:

By shifting the origin to a suitable point axes remaining parallel. The equation 5y² – 9x² + 30 y + 18 x – 9 = 0 reduces to form Y²/ b² – X²/ a² = 1 (a > 0 and b > 0). Find the coordinates of point at which the origin is shifted. Also find values of a and b.

Solution:

The old equation of locus is 5y² – 9x² + 30 y + 18 x – 9 = 0…………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

5(Y + k)² – 9(X + h)² + 3 (Y + k) + 18(X + h) – 9 = 0

∴   5(Y² + 2kY + k²) – 9(X²+ 2hX + h²) + 30Y + 30k + 18X + 18h – 9 = 0∴   5Y² + 10kY + 5k² – 9X² – 18hX – 9h² + 30Y + 30k + 18X + 18h – 9 = 0

∴   5Y² – 9X² + (10k + 30)Y + (- 18h + 18)X + (5k² – 9h² + 30k + 18h – 9) = 0 …………. (2)

The new equation of locus is Y²/ b² + X²/ a² = 1

Thus terms of X and Y are absent

∴   10k + 30 = 0 and – 18h + 18 = 0

∴   K = -3 and h = 1

∴   The origin is shifted to (1, -3)

The value of 5k² – 9h² + 30k + 18h – 9 = 5(-3) ² -9(1) ² + 30(-3) + 18(1) – 9

∴   5k² – 9h² + 30k + 18h – 9  =  45 – 9 – 90 + 18  – 9 = – 45

Equation (2) reduces to

5Y² – 9X² – 45 = 0

∴   5Y² – 9X² = 45

∴   5Y²/45 – 9X²/45 = 1

∴   Y²/9 – X²/5 = 1

Comparing with Y²/ b² – X²/ a² = 1

∴   b² = 9 and a² = 5

Hence b = 3 and a = √5 (only positive values are considered since a> o and b > 0)

Ans: The origin is shifted to  (1, -3) and a = √5 and b = 3

Example – 05:

By shifting the origin to O'(h, k) axes remaining parallel, reduce the equation 4x² + 9y² +16x – 18 y +24 = 0 reduces to form X²/ a² + Y²/ b² = 1 (a > 0 and b > 0). Find O'(h, k) at which the origin is shifted. Also find values of a and b.

Solution:

The old equation of locus is 4x² + 9y² +16x – 18 y +24 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

4(X + h)² + 9(Y + k)² +16(X + h) – 18(Y + k) +24 = 0

∴  4(X²+ 2hX + h²) + 9(Y² + 2kY + k²) +16(X + h) – 18(Y + k) +24 = 0

∴  4X²+ 8hX + 4h² + 9Y² + 18kY + 9k² +16X + 16h – 18Y – 18k +24 = 0

∴  4X² + 9Y²  + (8h + 16)X  + (18k – 18)Y + (4h² + 9k² + 16h – 18k +24) = 0 …………. (2)

The new equation of locus is X²/ a² + Y²/ b² = 1

Thus terms of X and Y are absent

∴  8h + 16 = 0 and 18k – 18 = 0

∴   h = -2 and k = 1

∴   The origin is shifted to O'(-2, 1)

The value of (4h² + 9k² + 16h – 18k +24) = 4(-2)² + 9(1)² + 16(-2) – 18(1) +24

∴  4h² + 9k² + 16h – 18k +24  =  16 + 9 -32 – 18 + 24 = – 1

Equation (2) reduces to

4X²+ 9Y² – 1 = 0

∴  4X²+ 9Y² = 1

∴  X²/(1/4) + Y²/(1/9) = 1

Comparing with X²/ a² + Y²/ b² = 1

∴   a² = 1/4 and b² = 1/9

Hence a = 1/2 and b = 1/3 (only positive values are considered since a> o and b > 0)

Ans: The origin is shifted to  O'(-2, 1) and a = 1/2 and b = 1/3

Example – 06:

By shifting the origin to suitable point axes remaining parallel, reduce the equation 2x² – y² – 4x + 4y – 3 = 0 reduces to form X²/ a² – Y²/ b² = 1 (a > 0 and b > 0). Find the coordinates of point at which the origin is shifted. Also find values of a and b.

Solution:

The old equation of locus is 2x² – y² – 4x + 4y – 3 = 0  …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

2(X + h)² – (Y + k)² – 4(X + h) + 4(Y + k) – 3 = 0

∴  2(X²+ 2hX + h²) – (Y² + 2kY + k²) – 4(X + h) + 4(Y + k) – 3 = 0

∴  2X²+ 4hX + 2h² – Y² – 2kY – k² – 4X – 4h + 4Y + 4k – 3 = 0

∴  2X² – Y² + (4h – 4)X + (-2k +4)Y + (2h² – k² – 4h + 4k – 3) = 0…………. (2)

The new equation of locus is X²/ a² – Y²/ b² = 1

Thus terms of X and Y are absent

∴  4h – 4 = 0 and -2k +4 = 0

∴   h = 1 and k = 2

∴   The origin is shifted to (1, 2)

The value of (2h² – k² – 4h + 4k – 3) = 2(1)² – (2)² – 4(1) + 4(2) – 3

∴  2h² – k² – 4h + 4k – 3  =  2 – 4 – 4 + 8 – 3 = -1

Equation (2) reduces to

2X²– Y² – 1 = 0

∴  2X²– Y² = 1

∴  X²/(1/2) + Y²/1 = 1

Comparing with X²/ a² + Y²/ b² = 1

∴   a² = 1/2 and b² = 1

Hence a = 1/√2 and b = 1 (only positive values are considered since a> o and b > 0)

Ans: The origin is shifted to  (1, 2) and a = 1/√2 and b = 1

Example – 07:

By shifting the origin to suitable point axes remaining parallel, the equation 3x² – 5y² –  6x – 20y – 32 = 0 reduces to form X²/ a² – Y²/ b² = 1 . Find the coordinates of the point at which the origin is shifted. Also find values of a and b.

Solution:

The old equation of locus is 3x² – 5y² –  6x – 20y – 32 = 0 …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

3(X + h)² – 5(Y + k)² –  6(X + h) – 20(Y + k) – 32 = 0

∴  3(X²+ 2hX + h²) – 5(Y² + 2kY + k²) – 6(X + h) – 20(Y + k) -32 = 0

∴  3X²+ 6hX + 3h² – 5Y² – 10kY – 5k² – 6X – 6h – 20Y – 20k -32 = 0

∴   3X² – 5Y² + (6h – 6)X + (- 10k – 20) Y + (3h²  – 5k² – 6h – 20k -32) = 0…………. (2)

The new equation of locus is X²/ a² – Y²/ b² = 1

Thus terms of X and Y are absent

∴  6h – 6 = 0 and – 10k – 20 = 0

∴   h = 1 and k = -2

∴   The origin is shifted to (1, -2)

The value of (3h²  – 5k² – 6h – 20k -32)  =3(1)²  – 5(-2)² – 6(1) – 20(-2) -32

∴  3h²  – 5k² – 6h – 20k -32  =  3  – 20 – 6 + 40 -32 = -15

Equation (2) reduces to

3X² – 5Y² – 15 = 0

∴  3X² – 5Y² = 15

∴  3X²/15 – 5Y²/15 = 1

∴  X²/5 – Y²/3 = 1

Comparing with X²/ a² + Y²/ b² = 1

∴   a² = 5 and a² = 3

Hence a = √5 and b = √3

Ans: The origin is shifted to  (1, -2) and a = √5 and b = √3

Example – 08:

By shifting the origin to suitable point axes remaining parallel, the equation y² –  6x + 4y + 28 = 0 reduces to form Y² = 4aX . Find the coordinates of the point at which the origin is shifted. Also, find the value of a

Solution:

The old equation of locus is y² –  6x + 4y + 28 = 0  …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

(Y + k)² –  6(X + h) + 4(Y + k) + 28 = 0

∴  Y² + 2kY + k² –  6X – 6h + 4Y + 4k + 28 = 0

∴  Y² + (2k + 4)Y + (k² – 6h + 4k + 28) = 6X …………. (2)

The new equation of locus is Y² = 4ax

Thus terms of Y and constant term are absent

∴  2k + 4 = 0 and k² – 6h + 4k + 28 = 0

∴   k = -2

and (-2)² – 6h + 4(-2) + 28 = 0

∴  4 – 6h – 8 + 28 = 0

∴   – 6h = – 24

∴  h = 4

∴   The origin is shifted to (4, -2)

Equation (2) reduces to

Y²  = 6x

Comparing with Y² = 4ax

∴   4a = 6 i.e. a = 3/2

Ans: The origin is shifted to  (4, -2) and a = 3/2

Example – 09:

By shifting the origin to suitable point axes remaining parallel, reduce the equation y² +  8x + 4y – 2 = 0 such that it will not contain the term in y and a constant term. Find the coordinates of the point at which the origin is shifted.

Solution:

The old equation of locus is y² +  8x + 4y – 2 = 0   …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k)

We have x = X + h and y = Y + k

Substituting these values in equation (1) we have

(Y + k)² + 8(X + h) + 4(Y + k) – 2 = 0

∴  Y² + 2kY + k² +  8X + 8h + 4Y + 4k – 2 = 0

∴  Y² + (2k + 4)Y + (k² + 8h + 4k – 2) + 8X = 0 …………. (2)

Thus terms of Y and constant term are absent

∴  2k + 4 = 0 and k² + 8h + 4k -2 = 0

∴   k = -2

and (-2)² + 8h + 4(-2) – 2 = 0

∴  4 + 8h – 8 – 2 = 0

∴   8h = 6

∴  h = 3/4

∴   The origin is shifted to (3/4, -2)

Ans: The origin is shifted to  (3/4, -2)

Example – 10:

The origin is shifted to point (3, p). Find the value of p so that the new equation of locus y2 +  2x – 8y + 7 = 0 such that it will not contain the term in y. Find the value of p.

Solution:

The old equation of locus is y² +  2x – 8y + 7 = 0    …………. (1)

Let (x, y) be the old coordinates and (X, Y) be the new coordinates after shifting the origin to (h, k) ≡ (3, p)

We have x = X + 3 and y = Y + p

Substituting these values in equation (1) we have

y² +  2x – 8y + 7 = 0

(Y + p)² + 2(X + 3)  – 8(Y + p) + 7 = 0

∴  Y² + 2pY + p² +  2X + 6 – 8Y – 8p + 7 = 0

∴  Y² + (2p – 8)Y + (p² – 8p + 13) + 2X = 0 …………. (2)

Thus terms of Y and constant term are absent

∴  2p – 8 = 0

∴   p = 4

Science > Mathematics > Coordinate Geometry > Locus > To Find Point at Which Origin is Shifted

The post To Find Point at Which Origin is Shifted appeared first on The Fact Factor.

Type – I: To Find New Coordinates After Shift of Origin

Example – 01:

If the origin is shifted to a point (1, 2), axes remaining parallel, find the new coordinates of the point (2, -3)

Solution:

New origin (1, 2) ≡ (h, k)

Let (x, y) ≡ (2, -3) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k

∴ X = 2 – 1 = 1 and Y = -3 -2 = -5

Ans: The new coordinates of point (2, -3) are (1, -5)

Example – 02:

If the origin is shifted to a point (1, 2), axes remaining parallel, find the new coordinates of the point (3, -4)

Solution:

New origin (1, 2) ≡ (h, k)

Let (x, y) ≡ (3, -4) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k

∴ X = 3 – 1 = 2 and Y = -4 -2 = -6

Ans: The new coordinates of point (3, -4) are (2, -6)

Example – 03:

If the origin is shifted to a point (2, -3), axes remaining parallel, find the new coordinates of the point (5, 2)

Solution:

New origin (2, -3) ≡ (h, k)

Let (x, y) ≡ (5, 2) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k

∴ X = 5 – 2 = 3 and Y = 2 + 3 = 5

Ans: The new coordinates of point (5, 2) are (3, 5)

Example – 04:

If the origin is shifted to a point (2, -3), axes remaining parallel, find the new coordinates of the point (2, -7)

Solution:

New origin (2, -3) ≡ (h, k)

Let (x, y) ≡ (2, -7) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k

∴ X = 2 – 2 = 0 and Y = -7 + 3 = -4

Ans: The new coordinates of point (2, -7) are (0, -4)

Example – 05:

If the origin is shifted to a point (2, -3), axes remaining parallel, find the new coordinates of the point (3, -4)

Solution:

New origin (2, -3) ≡ (h, k)

Let (x, y) ≡ (3, -4) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k

X = 3 – 2 = 1 and Y = -4 + 3 = -1

Ans: The new coordinates of point (3, -4) are (1, -1)

Example – 06:

If the origin is shifted to a point (2, -3), axes remaining parallel, find the new coordinates of the point (-10, 3)

Solution:

New origin (2, -3) ≡ (h, k)

Let (x, y) ≡ (-10, 3) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k

∴ X = -10 – 2 = -12 and Y = 3 + 3 = 6

Ans: The new coordinates of point (-10, 3) are (-12, 6)

Example – 07:

If the origin is shifted to a point (2, -3), axes remaining parallel, find the new coordinates of the point (2, -3)

Solution:

New origin (2, -3) ≡ (h, k)Let (x, y) ≡ (2, -3) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k

∴ X = 2 – 2 = 0 and Y = -3 + 3 = 0

Ans: The new coordinates of point (2, -3) are (0, 0)

Type – II: To Find Old Coordinates After Shift of Origin

Example – 08:

If the origin is shifted to a point (-3, 2), axes remaining parallel, the new coordinates of the point are (3, 1). Find the old coordinates.

Solution:

New origin (-3, 2) ≡ (h, k)

Let (X, Y) ≡ (3, 1) be the new coordinates and (x, y) be the old coordinates

We have x = X + h and y = Y + k

x = 3 -3 = 0 and Y = 1 + 2 = 3

Ans: The old coordinates of point (3, 1) are (0, 3)

Example – 09:

If the origin is shifted to a point (-3, 2), axes remaining parallel, the new coordinates of the point are (-5, -4). Find the old coordinates.

Solution:

New origin (-3, 2) ≡ (h, k)

Let (X, Y) ≡ (-5, -4) be the new coordinates and (x, y) be the old coordinates

We have x = X + h and y = Y + k

∴ x = -5 -3 = -8 and Y = -4 + 2 = -2

Ans: The old coordinates of point (-5, -4) are (-8, -2)

Example – 10:

If the origin is shifted to a point (-2, 1), axes remaining parallel, the new coordinates of the point are (2, 4). Find the old coordinates.

Solution:

New origin (-2, 1) ≡ (h, k)

Let (X, Y) ≡ (2, 4) be the new coordinates and (x, y) be the old coordinates

We have x = X + h and y = Y + k

∴ x = 2 -2 = 0 and Y = 4 + 1 = 5

Ans: The old coordinates of point (2, 4) are (0, 5)

Example – 11:

If the origin is shifted to a point (-2, 1), axes remaining parallel, the new coordinates of the point are (3, -5). Find the old coordinates.

Solution:

New origin (-2, 1) ≡ (h, k)

Let (X, Y) ≡ (3, -5) be the new coordinates and (x, y) be the old coordinates

We have x = X + h and y = Y + k

∴ x = 3 -2 = 1 and Y = -5 + 1 = -4

Ans: The old coordinates of point (3, -5) are (1, -4)

Example – 12:

If the origin is shifted to a point (-2, 1), axes remaining parallel, the new coordinates of the point are (0, 4). Find the old coordinates.

Solution:

New origin (-2, 1) ≡ (h, k)

Let (X, Y) ≡ (0, 4) be the new coordinates and (x, y) be the old coordinates

We have x = X + h and y = Y + k

x = 0 -2 = -2 and Y = 4 + 1 = 5

Ans: The old coordinates of point (0, 4) are (-2, 5)

Example – 13:

If the origin is shifted to a point (-2, 1), axes remaining parallel, the new coordinates of the point are (-4, 8). Find the old coordinates.

Solution:

New origin (-2, 1) ≡ (h, k)

Let (X, Y) ≡ (-4, 8) be the new coordinates and (x, y) be the old coordinates

We have x = X + h and y = Y + k

∴ x = -4 -2 = -6 and Y = 8 + 1 = 9

Ans: The old coordinates of point (-4, 8) are (-6, 9)

Example – 14:

If the origin is shifted to a point (-2, 1), axes remaining parallel, the new coordinates of the point are (-6, 3). Find the old coordinates.

Solution:

New origin (-2, 1) ≡ (h, k)

Let (X, Y) ≡ (-6, 3) be the new coordinates and (x, y) be the old coordinates

We have x = X + h and y = Y + k

x = -6 -2 = -8 and Y = 3 + 1 = 4

Ans: The old coordinates of point (-6, 3) are (-8, 4)

Type – III: Find Coordinates of Point Where Origin is Shifted (Location of Shift of Origin)

Example – 15:

The point (3, 8) becomes (-2, 1) after the shift of origin. Find the coordinates of the point, where the origin is shifted.

Solution:

Let the new origin be (h, k)

Let (x, y) ≡ (3, 8) be the old coordinates and (X, Y) ≡ (-2, 1) be the new coordinates

We have h = x – X and k = y – Y

∴ h = 3 + 2 = 5 and Y = 8 – 1 = 7

Ans: The coordinates of the point, where the origin is shifted are (5, 7)

Example – 16:

The point (2, 3) becomes (5, -2) after the shift of origin. Find the coordinates of the point, where the origin is shifted.

Solution:

Let the new origin be (h, k)

Let (x, y) ≡ (2, 3) be the old coordinates and (X, Y) ≡ (5, -2) be the new coordinates

We have h = x – X and k = y – Y

∴ h = 2 – 5 = -3 and Y = 3 + 2 = 5

Ans: The coordinates of the point, where the origin is shifted are (-3, 5)

Example – 17:

The point (3, -4) becomes (7, 2) after the shift of origin. Find the coordinates of the point, where the origin is shifted.

Solution:

Let the new origin be (h, k)

Let (x, y) ≡ (3, -4) be the old coordinates and (X, Y) ≡ (7, 2) be the new coordinates

We have h = x – X and k = y – Y

∴ h = 3 – 7 = -4 and Y = -4 -2 = -6

Ans: The coordinates of the point, where the origin is shifted are (-4, -6)

Example – 18:

The point (a, b) becomes (c, d) after the shift of origin. Find the coordinates of the point, where the origin is shifted.

Solution:

Let the new origin be (h, k)

Let (x, y) ≡ (a, b) be the old coordinates and (X, Y) ≡ (c, d) be the new coordinates

We have h = x – X and k = y – Y

∴ h = a – c and Y = b – d

Hence the coordinates of the point, where the origin is shifted are (a – c, b – d)

Miscellaneous:

Example – 19:

The origin is shifted to the point (-5, 9)axes remaining parallel if the point (3, b) lies on the new x-axis and point (a, 3) lies on the new y-axis, find the values of a and b

Solution:

New origin (-5, 9) ≡ (h, k)

Let (x, y) ≡ B(3, b) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k

∴ X = 3 + 5 = 8 and Y = b – 9

Hence new coordinates of point B are (8, b- 9)

Now point B lies on new x-axis i.e. Y = 0

b – 9 = 0

b = 9

Let (x, y) ≡ A(a, 3) be the old coordinates and (X, Y) be the new coordinates

We have X = x – h and Y = y – k∴ X = a + 5  and Y = 3 – 9 = – 6

Hence new coordinates of point A are (a + 5, – 6)

Now point B lies on new y-axis i.e. X = 0

a + 5 = 0

a = -5

Ans: a = -5 and b = 9

Science > Mathematics > Coordinate Geometry > Locus > Shift of Origin

The post Shift of Origin appeared first on The Fact Factor.

In this article, we shall study the concept of locus and to find the equation of the locus.

Example – 46:

A(3, 1) and B(2, -5) are the vertices of a triangle. Find the equation of locus of the third vertex C, if the centroid of the triangle lies on the locus y = 3 + 2x².

Solution:

let G(h, k) be the centroid. It lies on the locus y = 3 + 2x²

Hence it should satisfy the equation of the locus

k = 3 + 2h²  ……….. (1)

Let C(x’, y’) be the third vertex of the triangle, whose locus is to be found

By Centroid formula

Substituting in equation (1)

(-4 + y’)/3 = 3 + 2((5+x’)/3)²

(-4 + y’)/3 = 3 + 2(25+ 10x’ +x’²)/9

Multiplying both sides by 9

3(-4 + y’) = 27 + 2(25+ 10x’ +2x’²)

-12 + 3y’ = 27 + 50+ 20x’ + 2x’²

27 + 50+ 20x’ + 2x’² + 12 – 3y’ = 0

2x’² + 20x’ – 3y’ + 89 = 0

Replacing x’ with x and y’ with y we get

2x² + 20x – 3y + 89 = 0

Ans: The equation of the required locus is 2x² + 20x – 3y + 89 = 0

Example – 47:

A(3, -2) and B(-2, 4) are the vertices of a triangle. Find the equation of locus of the third vertex C, if the centroid of the triangle lies on the locus 3x² – 5y² = 15.

Solution:

let G(h, k) be the centroid. It lies on the locus 3x² – 5y² = 15

Hence it should satisfy the equation of the locus.

3h² – 5k² = 15 ……….. (1)

Let C(x’, y’) be the third vertex of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

3((1+ x’)/3)² – 5((2 + y’)/3)² = 15

3((1+ 2x’ + x’²)/9) – 5((4 + 4y’ + y’²)/9) = 15

Multiplying both sides by 9

3(1+ 2x’ + x’²) – 5(4+ 4y’ + y’²) = 135

3+ 6x’ + 3x’² – 20 – 20y’ – 5y’²  – 135 = 0

3x’² – 5y’²  + 6x’ – 20y’ – 152 = 0

Replacing x’ with x and y’ with y we get

3x² – 5y²  + 6x – 20y – 152 = 0

Ans: The equation of the required locus is 3x² – 5y²  + 6x – 20y – 152 = 0

Example – 48:

A(2, -5) and B(-7, 6) are the vertices of a triangle. Find the equation of locus of the third vertex C, if the centroid of the triangle lies on the locus 3x – 4y = 11.

Solution:

let G(h, k) be the centroid. It lies on the locus 3x – 4y + 11 = 0

Hence it should satisfy the equation of the locus.

3h – 4k + 11 = 0 ……….. (1)

Let C(x’, y’) be the third vertex of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

3((-5+ x’)/3) – 4((1 + y’)/3) + 11 = 0

Multiplying both sides by 3

3(-5+ x’) – 4(1 + y’) + 33 = 0

15+ 3x’ – 4 – 4y’ + 33 = 0

3x’ – 4y’ – 19 + 33 = 0

3x’ – 4y’ + 14 = 0

Replacing x’ with x and y’ with y we get

3x – 4y + 14 = 0

Ans: The equation of the required locus is 3x – 4y + 14 = 0

Example – 49:

A(5, -2) and B(2, 4) are the vertices of a triangle. The third vertex C of triangle lies on the locus y = 1 + x + x². Find the equation of locus of the centroid of the triangle.

Solution:

let C(h, k) be the centroid. It lies on the locus y = 1 + x + x²

Hence it should satisfy the equation of the locus.

k = 1 + h + h² ……….. (1)

Let G(x’, y’) be the centroid of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

(3y’ – 2) = 1 + (3x’ – 7) + (3x’ – 7)²

3y’ – 2 = 1 + 3x’ – 7 + 9x’² – 42x’ + 49

1 + 3x’ – 7 + 9x’² – 42x’ + 49 – 3y’ + 2 = 0

9x’² – 39x’ – 3y’ + 45 = 0

3x’² – 13x’ – y’ + 15 = 0

Replacing x’ with x and y’ with y we get

3x² – 13x – y + 15 = 0

Ans: The equation of the required locus is 3x² – 13x – y + 15 = 0

Example – 50:

A(3, -2) and B(-2, 4) are the vertices of a triangle. The third vertex C of triangle lies on the locus y = 2 + 3x + x². Find the equation of locus of the centroid of the triangle.

Solution:

let C(h, k) be the centroid. It lies on the locus y = 2 + 3x + x²

Hence it should satisfy the equation of the locus.

k = 2 + 3h + h² ……….. (1)

Let G(x’, y’) be the centroid of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

3y’ – 7 = 2 + 3(3x’ – 2) + (3x’ – 2)²

3y’ – 7 = 2 + 9x’ – 6 + (9x’² – 12x’ + 4)

2 + 9x’ – 6 + 9x’² – 12x’ + 4 – 3y’ + 7 = 0

9x’² – 3x’ – 3y’ + 7 = 0

Replacing x’ with x and y’ with y we get

9x² – 3x – 3y + 7 = 0

Ans: The equation of the required locus is 9x² – 3x – 3y + 7 = 0

Example – 51:

A(1, 4) and B(6, -3) are the vertices of a triangle. The third vertex C of the triangle lies on the locus 5x + 4y = 9. Find the equation of locus of the centroid of the triangle.

Solution:

let C(h, k) be the centroid. It lies on the locus 5x + 4y = 9

Hence it should satisfy the equation of the locus.

5h + 4k = 9 ……….. (1)

Let G(x’, y’) be the centroid of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

5(3x’ – 7) + 4(3y’ – 1) = 9

15x’ – 35 + 12y’ – 4 = 9

15x’ + 12y’ -39 – 9 = 0

15x’ + 12y’ – 48 = 0

5x’ + 4y’ – 16 = 0

Replacing x’ with x and y’ with y we get

5x + 4y – 16 = 0

Ans: The equation of the required locus is 5x + 4y – 16 = 0

Example – 52:

A(4, -3) and C(0, 2) are the vertices of a triangle. The third vertex B of the triangle lies on the locus y = 1+ x². Find the equation of locus of the centroid of the triangle.

Solution:

let B(h, k) be the centroid. It lies on the locus y = 1+ x²

Hence it should satisfy the equation of the locus.

k = 1+ h² ……….. (1)

Let G(x’, y’) be the centroid of the triangle, whose locus is to be found.

By Centroid formula

Substituting in equation (1)

(3y’ + 1) = 1 +(3x’ – 4)²

3y’ + 1 = 1 +9x’² – 24x’ + 16

1 +9x’² – 24x’ + 16 – 3y’ – 1 = 0

9x’² – 24x’  – 3y’ + 16 = 0

Replacing x’ with x and y’ with y we get

9x² – 24x  – 3y + 16 = 0

Ans: The equation of the required locus is 9x² – 24x  – 3y + 16 = 0

Science > Mathematics > Coordinate Geometry > Locus > Equation of the Locus

The post More Problems on Equation of Locus: Set – III appeared first on The Fact Factor.

In this article, we shall study the method to check whether a given point lies on the locus or not and to find a point on locus satisfying the given condition.

To Check Whether the Point Lies on the Locus:

Algorithm:

1. Write the equation of given locus
2. To check whether the point (x₁, y₁) lies on the locus, substitute x = x₁ and y = y₁ in L.H.S. of the equation of locus.
3. If L.H.S.. = R.H.S., the point lies on the locus, and If L.H.S. ≠ R.H.S., the point does not lie on the locus.

Example – 01:

Examine whether points (4, -1) and (3, 4) lies on the locus 2x² + 2y² – 5x + 11y – 3 = 0.

Solution:

To check point (4, -1)

The equation of the locus is 2x² + 2y² – 5x + 11y – 3 = 0

Substituting x = 4 and y = -1, in L.H.S. of equation of locus

L.H.S. = 2(4)² + 2(-1)² – 5(4) + 11(-1) – 3

∴  L.H.S. = 32 + 2 – 20  – 11 – 3 = 0 = R.H.S.

Hence point (4, -1) lies on given locus

To check point (3,4)

The equation of the locus is 2x² + 2y² – 5x + 11y – 3 = 0

Substituting x = 3 and y = 4, in L.H.S. of equation of locus

L.H.S. = 2(3)² + 2(4)² – 5(3) + 11(4) – 3

∴  L.H.S. = 18 + 32 – 15  + 44 – 3 = 76 ≠ R.H.S.

Hence point (3, 4) does not lie on given locus

Example – 02:

Examine whether points (5, 2), (2, 3) and (1, -4) lies on the locus x² + y² + 6x – 6y – 47 = 0.

Solution:

To check point (5, 2)

The equation of the locus is x² + y² + 6x – 6y – 47 = 0

Substituting x = 5 and y = 2, in L.H.S. of equation of locus

L.H.S. = (5)² + (2)² + 6(5) – 6(2) – 47

∴  L.H.S. = 25 + 4 + 30  – 12 – 47 = 0 = R.H.S.

Hence point (5, 2) lies on given locus

To check point (2, 3)

The equation of the locus is x² + y² + 6x – 6y – 47 = 0

Substituting x = 2 and y = 3, in L.H.S. of equation of locus

L.H.S. = (2)² + (3)² + 6(2) – 6(3) – 47

∴  L.H.S. = 4 + 9 + 12  – 18 – 47 = – 40 ≠ R.H.S.

Hence point (2, 3) does not lie on given locus

To check point (1, -4)

The equation of the locus is x² + y² + 6x – 6y – 47 = 0

Substituting x = 2 and y = 3, in L.H.S. of equation of locus

L.H.S. = (1)² + (-4)² + 6(1) – 6(-4) – 47

∴  L.H.S. = 1 + 16 + 6  + 24 – 47 = 0 = R.H.S.

Hence point (1, -4) lies on given locus

Example – 03:

Show that the point P(at², 2at) lies on the locus  y² = 4ax

Solution:

The equation of the locus is y² = 4ax

Substituting x = at²and y = 2at, in L.H.S. of equation of locus

L.H.S. = y² = (2at)² = 4a²t²   ………. (1)

∴  R.H.S. = 4ax = 4a (at²) = 4a²t²   ………. (2)

From equations (1) and (2) we have

L.H.S. = R.H.S.

Hence point P(at², 2at) lies on the locus y² = 4ax

Example – 04:

Show that the point Q(a cosθ, b sinθ) lies on the locus  x²/a² + y²/b² = 1

Solution:

The equation of the locus is x²/a² + y²/b² = 1

Substituting x = a cosθand y = b sinθ, in L.H.S. of equation of locus

L.H.S. = (a cosθ)²/a² + (b sinθ)²/b²

L.H.S. = a² cos²θ/a² + b² sin²θ/b² = cos²θ + sin²θ = 1 = R.H.S..

Hence point Q(a cosθ, b sinθ) lies on the locus x²/a² + y²/b² = 1

Example – 05:

Show that the point R(a secθ, b tanθ) lies on the locus  x²/a² – y²/b² = 1

Solution:

The equation of the locus is x²/a² – y²/b² = 1

Substituting x = a secθand y = b tanθ, in L.H.S. of equation of locus

L.H.S. = (a secsθ)²/a² – (b tanθ)²/b²

L.H.S. = a² sec²θ/a² – b² tan²θ/b² = sec²θ – tan²θ = 1 = R.H.S.

Hence point R(a secθ, b tanθ) lies on the locus x²/a² – y²/b² = 1

Example – 06:

Show that the point S(a cosθ, a sinθ) lies on the locus  x² + y² = a²

Solution:

The equation of the locus is x² + y² = a²

Substituting x = a cosθand y = a sinθ, in L.H.S. of equation of locus

L.H.S. = (a cosθ)² + (a sinθ)²

L.H.S. = a² cos²θ + a² sin²= a²(cos²θ + sin²θ) = a²(1) = a² =R.H.S.

Hence point S(a cosθ, a sinθ) lies on the locus x² + y² = a²

Example – 07:

Show that the point T(5 cosθ, 5 sinθ) lies on the locus  x² + y² = 25

Solution:

The equation of the locus is x² + y² = 25

Substituting x = 5 cosθand y = 5 sinθ, in L.H.S. of equation of locus

L.H.S. = (5 cosθ)² + (5 sinθ)²

L.H.S. = 25 cos²θ + 25 sin²= 25(cos²θ + sin²θ) = 25(1) = 25 =R.H.S.

Hence point T(5 cosθ, 5 sinθ) lies on the locus x² + y² = 25

Ans: k = -9 and a = – 4/3

Example – 08:

show that for all values of r, the point (x₁ + r cosθ, y₁ + r sinθ) always lies on locus y – y₁ = tanθ (x – x₁)

Solution:

The equation of the locus is y – y₁ = tanθ (x – x₁)

Point (x₁ + r cosθ, y₁ + r sinθ) lies on it

Substituting x = x₁ + r cosθ and y₁ + r sinθ

L.H.S. = (y₁ + r sinθ) – y₁ = r sinθ  …….. (1)

R.H.S. = tanθ ((x₁ + r cosθ) – x₁) = tanθ (r cosθ)

∴  RH.S. = (sinθ/cosθ). r cosθ = r sinθ  …….. (2)

From equations (1) and (2) for all values of r we have

L.H.S. = R.H.S.

Hence point (x₁ + r cosθ, y₁ + r sinθ)  lies on the locus y – y₁ = tanθ (x – x₁)

To Find Value of Arbitrary Constant When Point on Locus is Given:

Algorithm:

1. Write the equation of given locus
2. Let given point be (x₁, y₁) lies on the locus, substitute x = x₁ and y = y₁ in the equation of locus.
3. solve an equation to find the value of constant.

Example – 09:

Point (-6, 3) lies on the locus x² = 4ay. Find the value of ‘a’.

Solution:

The equation of the locus is x² = 4ay

Point (-6, 3) lies on it

Substituting x = – 6 and y = 3, in the equation of locus

(-6)² = 4a(3)

∴  36 = 12 a

∴  a = 3

Example – 10:

Points (3, 2) and (-1, -2) lie on locus ax + by = 5 . Find the value of ‘a’ and ‘b’.

Solution:

The equation of the locus is ax + by = 5

Point (3, 2) lies on it

Substituting x = 3 and y = 2, in the equation of locus

a(3) + b(2) = 5

∴  3a + 2b = 5    …………. (1)

Point (-1, -2) lies on it

Substituting x = -1 and y = -2, in the equation of locus

a(-1) + b(-2) = 5

∴  a + 2b = -5    …………. (2)

Solving (1) and (2) simultaneously

a = 5 and b = – 5

Example – 11:

Points (-4, 4) and (-16, b) lie on locus y² = ax. Find the value of ‘a’ and ‘b’.

Solution:

The equation of the locus is y² = ax

Point (-4, 4) lies on it

Substituting x = -4 and y = 4, in the equation of locus

(4)² = a(-4)

∴  16 = – 4a

∴  a = -4

Point (-16, b) lies on it

Substituting x = -16 and y = b, in the equation of locus y² = ax

b² = (- 4)(- 16) = 64

∴  b = ± 8

Ans:  a = -4 and b = ± 8

Example – 12:

Point (-8, 6) lies on locus x²/4 + y²/3 = k . Find the value of ‘k’.

Solution:

The equation of the locus is x²/4 + y²/3 = k

Point (-8, 6) lies on it

Substituting x = -8 and y = 6, in the equation of locus

(-8)²/4 + (6)²/3 = k

∴  64/4 + 36/3 = k

∴  16 + 12 = k

∴  k = 28

Example – 13:

Points P(-2, 2) and Q(3, a) lie on locus x² – 7x + ky = 0. Find the values of ‘k’ and ‘a’.

Solution:

The equation of the locus is x² – 7x + ky = 0

Point (-2, 2) lies on it

Substituting x = -2 and y = 2, in the equation of locus

(-2)² – 7(-2) + k(2) = 0

∴  4 + 14 + 2k = 0

∴  2 k = -18

∴  k = – 9

Point (3, a) lies on it

Substituting x = -2 and y = 2, in the equation of locus

(3)² – 7(3) + k(a) = 0

∴  9 – 21+ (-9)a= 0

∴  – 9a = 12

∴  a = 12/-9 = – 4/3

To Find Point of Intersection Where the Locus Cuts x-axis and y- axis:

Algorithm to Find Point of Intersection:

1. To find the  point of intersection (a, 0) on x-axis put x = a and y = 0, in the equation of locus. Find the value of a, then (a, 0) is the point of intersection of the locus with the x-axis.
2. To find the point of intersection on y-axis put x = 0 and y = b in the equation of locus. Find the value of b, then (0, b) is the point of intersection of the locus with the y-axis.

Example – 14:

Find the point of intersection of the locus x² + y² – 2x – 14y -15 = 0 with a) x- axis and b) y-axis

Solution:

Equation of the locus is x² + y² – 2x – 14y -15 = 0

Let (a, 0) be the point where the locus cuts x-axis

substituting x = a and y = 0 in equation of locus

a² + 0² – 2a – 14(0) -15 = 0

∴   a² – 2a  -15 = 0

∴   (a – 5)(a +3) = 0

∴   a = 5 or a = -3

Thus point of intersection of locus with x- axis are (5, 0) and (-3, 0)

Let (0, b) be the point where the locus cuts y-axis

substituting x = 0 and y = b in equation of locus

0² + b² – 2(0) – 14(b) -15 = 0

∴   b² – 14 b  -15 = 0

∴   (b – 15)(b + 1) = 0

∴   b = 15 or b = -1

Thus point of intersection of locus with y- axis are (0. 15) and (0, -1)

Example – 15:

Find the point of intersection of the locus x² + y² – 4x – 6y -12 = 0 with the x-axis and find the length of intercept made by the curve on the x-axis.

Solution:

Equation of the locus is x² + y² – 4x – 6y -12 = 0

Let (a, 0) be the point where the locus cuts x-axis

substituting x = a and y = 0 in equation of locus

a² + 0² – 4a – 6(0) -12 = 0

∴   a² – 4a  -12 = 0

∴   (a – 6)(a +2) = 0

∴   a = 6 or a = -2

Thus point of intersection of locus with x- axis are (6, 0) and (-2, 0)

Let A(6, 0) and (-2, 0) be the points

∴ the length of intercept made by the curve on x-axis = AB

AB² = (x₁ – x₂)² + (y₁ – y₂)²

∴  AB² = (6 + 2)² + (0 – 0)² = 64

∴  AB = 8

Thus points of intersection of the locus with x-axis are (6, 0) and (-2, 0) and the length of intercept made by the curve on x-axis = 8 unit

Note:

As both the points are on x-axis AB = |Difference in x coordinates| = |6 – (-2)| = 8 unit

Example – 16:

Find the point of intersection of the locus 16x² + 25y² = 400 with the x-axis and find the length of intercept made by the curve on the x-axis.

Solution:

Equation of the locus is 16x² + 25y² = 400

Let (a, 0) be the point where the locus cuts x-axis

substituting x = a and y = 0 in equation of locus

16a² + 25(0)² = 400

∴  16a²  = 400

∴  a²  = 400/16

∴   a =± 20/4  =± 5

Thus points of intersection of locus with x- axis are (5, 0) and (-5, 0)

Let A(5, 0) and (-5, 0) be the points

∴ the length of intercept made by the curve on x-axis = AB

AB² = (x₁ – x₂)² + (y₁ – y₂)²

∴  AB² = (5 + 5)² + (0 – 0)² = 100

∴  AB = 10

Thus points of intersection of locus with x- axis are (5, 0) and (-5, 0)

and the length of intercept made by the curve on x-axis = 10 unit

Note: 

As both the point are on x-axis AB = |Difference in x coordinates| = |5 – (-5)| = 10 unit

Example – 17:

Find the point on y-axis which also lie on the locus 3x² – 5xy + 6y² – 54 = 0.

Solution:

Equation of the locus is 3x² – 5xy + 6y² – 54 = 0

Let (0, b) be the point where the locus cuts y-axis

substituting x = 0 and y = b in equation of locus

3(0)² – 5(0)(b) + 6(b)² – 54 = 0

∴  6b²  = 54

∴  b²  = 9

∴   b =± 3

Thus point of intersection of locus with y- axis are (0, 3) and (0, -3)

Science > Mathematics > Coordinate Geometry > Locus > Point on Locus

The post Point on Locus appeared first on The Fact Factor.

In this article we shall study the concept of locus and to find the equation of locus of a point using distance formula.

Example – 31:

Find the equation of the locus of a point such that its distance from (5, 0) is equal to a distance from the y-axis.

Solution:

Let P(x. y) be the point on the locus, A(5. 0) be the point

Distance of point from y-axis = x

Given: AP = x

∴ AP² = x²

∴ [(x – 5 )²  + (y – 0)² ] = x²

∴ x ²  – 10x + 25 + y ²  = x²∴ – 10x + 25 + y ²  = 0

∴   y ² – 10x + 25   =0

Hence required equation of the locus  is  y ² – 10x + 25   = 0

Example – 32:

Find the equation of the locus of a point which is equidistant from (1, 3) and the x-axis.

Solution:

Let P(x. y) be the point on the locus, A(1, 3) be the point

Distance of point from x-axis = y

Given: AP = y

∴ AP² = y²

∴ [(x – 1 )²  + (y – 3)² ] = y²

∴ x ²  – 2x + 1 + y ² – 6y + 9  = y²

∴ x ²  – 2x  – 6y + 10  = 0

Hence required equation of the locus  is x ²  – 2x  – 6y + 10  = 0

Example – 33:

Find the equation of locus of a point such that, the sum of the square of its distances from the points (2, 5) and (3, -1) is 40.

Solution:

Let P(x. y) be the point on the locus, Given A(2, 5) and B(3, -1) be the given points

Given  PA² + PB² = 40

∴ [(x – 2 )²  + (y – 5)² ] +  [(x – 3)²  + (y + 1)²] = 40

∴ x²  – 4x + 4 + y² – 10y + 25 + x² – 6x + 9  + y² + 2y + 1 = 40

∴ 2x² + 2y²  – 10x   – 8y – 1 = 0

Hence required equation of the locus  is 2x² + 2y²  – 10x   – 8y – 1 = 0

Example – 34:

Find the equation of locus of a point such that, its distance from (a, 0) is m times  its distance from (0, a)

Solution:

Let P(x. y) be the point on the locus, Given A(a, o) and B(0, a) be the given points

Given  PA = m. PB

∴   PA² = m². PB²

∴ [(x – a )²  + (y – 0)² ] =  m²[(x – 0)²  + (y – a)²]

∴ x²  – 2ax + a² + y² = m²( x² + y² – 2ay + a²)

∴ x²  – 2ax + a² + y² = m²x² + m²y² – 2am²y + a²m²

∴ x²  – 2ax + a² + y² – m²x² – m²y² + 2am²y – a²m² = 0

∴ (1 – m²) x²  + ( 1 – m²) y²  – 2ax  + 2am²y +  (1 – m²) a² = 0

∴ (1 – m²)( x²  +  y² ) – 2ax  + 2am²y +  a²(1 – m²) = 0

Hence required equation of the locus  is (1 – m²)( x²  +  y² ) – 2ax  + 2am²y +  a²(1 – m²) = 0

Example – 35:

The point S is (3, 0) and abscissa of point M is -3. A variable point P is such that ordinates of P and M are equal. Find the equation of locus of point P such that SP = PM.

Solution:

Let P(x. y) be the point on the locus, Given S(3,0)

The abscissa of point M is -3 and its ordinate is same as P i.e. y. Hence M(-3, y) is the point

Given  SP = PM

∴  SP² = PM²

∴ [(x – 3 )²  + (y – 0)² ] =  [(x + 3)²  + (y – y)²]

∴ x²  – 6x + 9 + y² =  x² + 6x + 9  + 0

∴ x²  – 6x + 9 + y² –  x² – 6x – 9  = 0

∴ – 12x   + y² = 0

∴  y² – 12x = 0

Hence required equation of the locus  is y² – 12x = 0

Example – 36:

Find the equation of locus of a point which moves such that the ratio of its distances from (2, 0) and (1, 3) is 5:4.

Solution:

Let P(x. y) be the point on the locus and A(2,0) and B(1, 3) be the points

Given  PA/ PB = 5/4

∴  4 PA = 5 PB

∴  16 PA² = 25 PB²

∴ 16 [(x – 2)²  + (y – 0)² ] =  25[(x – 1)²  + (y – 3)²]

∴ 16(x²  – 4x + 4 + y² ) =  25(x² – 2x + 1  + y² – 6y + 9)

∴ 16x²  – 64x + 64 + 16y²  =  25x² – 50x + 25  + 25 y² – 150y + 225

∴  25x² – 50x + 25  + 25 y² – 150y + 225 – 16x²  + 64x – 64 – 16y²  = 0

∴  9x² + 9 y²  + 14x  – 150y – 186  = 0

Hence required equation of the locus  is 9x² + 9 y²  + 14x  – 150y – 186  = 0

Example – 37:

Find the equation of locus of a point such that the sum of its distances from co-ordinate axes is thrice its distance from the origin.

Solution:

Let P(x. y) be the point on the locus and O(0,0) be the origin

Distance of point from y-axis = x

Distance of point from x-axis = y

Given  x + y = 3OP

(x + y)² = 9OP²

∴ x² + 2xy + y²=  9[(x – 0)²  + (y – 0)²]

∴ x² + 2xy + y² =  9x ²  + 9y²

∴   9x ²  + 9y² – x² – 2xy – y² = 0

∴   8x ²  + 8y² – 2xy  = 0

∴   4x ²  + 4y² – xy  = 0

Hence required equation of the locus  is 4x ² – xy + 4y²= 0

Example – 38:

A(4, 0) and B(-4, 0) are two given points. A variable point P such that PA + PB = 10, show that the equation of locus of point P is

Solution:

Let P(x. y) be the point on the locus and A(4, 0) and B(-4, 0) be the points

Given  PA  + PB = 10

∴  √(x – 4)²  + ( y – 0)²   + √(x + 4)²  + ( y – 0)²   =10

∴  √x² – 8x + 16  + y ²   + √x² + 8x + 16  + y²   =10

∴  √x² – 8x + 16  + y ²   = 10 – √x² + 8x + 16  + y²  

Squaring both sides

∴  x² – 8x + 16  + y ² = 100 – 20 √x² + 8x + 16  + y²  + x² + 8x + 16  + y ²

∴  – 8x  = 100 – 20 √x² + 8x + 16  + y²   + 8x

∴  20 √x² + 8x + 16  + y²   =100 + 16x

Squaring both sides

400(x² + 8x + 16  + y ²) = 10000 + 3200x + 256x²

∴   400x² + 3200x + 6400  + 400y ² = 10000 + 3200x + 256x²

∴   400x² – 256x² + 400y ² = 10000  – 6400

∴   144x² + 400y ² = 3600

Dividing both sides by 3600

Proved as required

Example – 39:

A(5, 0) and B(-5, 0) are two given points. A variable point P such that PA – PB = 6, show that the equation of locus of point P is

Solution:

Let P(x. y) be the point on the locus and A(5, 0) and B(-5, 0) be the points

Given  PA  –  PB = 6

∴  √(x – 5)²  + ( y – 0)²   – √(x + 5)²  + ( y – 0)²   =6

∴  √x² – 10x + 25  + y ²   – √x² + 10x + 25  + y²   = 6

∴  √x² – 10x + 25  + y ²   = 6 + √x² + 10x + 25  + y²  

Squaring both sides

∴  x² – 10x + 25  + y ² = 36 + 12 √x² + 10x + 25  + y²  + x² + 10x + 25  + y ²

∴  – 10x  = 36 + 12 √x² + 10x + 25  + y²   + 10x

∴  – 12 √x² + 10x + 25  + y²   = 36 + 20x

Squaring both sides

144(x² + 10x + 25 + y ²) = 1296 + 1440x + 400x²

∴  144x² + 1440x + 3600 + 144 y ² = 1296 + 1440x + 400x²

∴   400x² – 144x² – 144 y ² = 3600 – 1296

∴   256x² – 144 y ² = 2304

Dividing both sides by 2304

Proved as required

Example – 40:

A(-3, 0) and B(3, 0) are two given points. A variable point P such that AP – PB = 10, show that the equation of locus of point P is

Solution:

Let P(x. y) be the point on the locus and A(-3, 0) and B(3, 0) be the points

Given  AP – PB = 10

∴  √(x + 3)²  + ( y – 0)²   – √(x – 3)²  + ( y – 0)²   =10

∴  √x² + 6x + 9  + y ²   – √x² – 6x + 9  + y²   =10

∴  √x² +  6x + 9  + y ²   = 10 – √x² – 6x + 9  + y²  

Squaring both sides

∴  x² + 6x + 9  + y ² = 100 – 20 √x² – 6x + 9  + y²  + x² – 6x + 9  + y ²

∴  6x  = 100 – 20 √x² – 6x + 9  + y²   – 6x

∴  20 √x² – 6x + 9  + y²   =100 – 12x

Squaring both sides

400(x² – 6x + 9  + y ²) = 10000 – 2400x + 144x²

∴  400x² – 2400x + 3600  + 400y ² = 10000 – 2400x + 144x²

∴   400x² – 144x² + 400y ² = 10000  – 3600

∴   256x² + 400y ² = 6400

Dividing both sides by 6400

Proved as required

Example – 41: 

A(2, 0) and B(-2, 0) are two given points. A variable point P such that the sum of its distances from given points is 6. Find the equation of the locus of point P.

Solution:

Let P(x. y) be the point on the locus and A(2, 0) and B(-2, 0) be the points

Given  PA  + PB = 10

∴  √(x – 2)²  + ( y – 0)²   + √(x + 2)²  + ( y – 0)²   = 6

∴  √x² – 4x + 4  + y ²   + √x² + 4x + 4  + y²   = 6

∴  √x² – 4x + 4  + y ²   = 6 – √x² + 4x + 4  + y²  

Squaring both sides

∴  x² – 4x + 4  + y ² = 36 – 12 √x² + 4x + 4  + y²  + x² + 4x + 4  + y ²

∴  – 4x  = 36 – 12 √x² + 4x + 4  + y²   + 4x

∴  12 √x² + 4x + 4  + y²   =36 + 8x

Squaring both sides

144(x² + 4x + 4  + y ²) = 1296 + 576x + 64x²

∴  144x² +  576x + 576  + 144y ² = 1296 + 576x + 64x²

∴  144x²  + 144y ² – 64x² = 1296 – 576

∴  80x²  + 144y ²= 720

Dividing both sides by 720

This is the required equation of the locus.

Example – 42:

A(c, 0) and B(-c, 0) are two given points. A variable point P such that the sum of its distances from the given point is 2a. Find the equation of locus if b² = a² – c².

Solution:

Let P(x. y) be the point on the locus and A(c, 0) and B(-c, 0) be the points

Given  AP – PB = 10

∴  √(x – c)²  + ( y – 0)²   – √(x + c)²  + ( y – 0)²   = 2a

∴  √x² – 2cx + c²  + y ²   – √x²+ 2cx + c² + y²   = 2a

∴  √x² – 2cx + c²  + y ²   = 2a – √x² + 2cx + c²  + y² 

Squaring both sides

∴  x² – 2cx + c ²  + y ² = 4a ² – 4a √x² + 2cx + c²  + y²  + x² + 2cx + c²  + y ²

∴  – 2cx  = 2a – 4a √x² + 2cx + c²  + y²    + 2cx

∴  4a √x² + 2cx + c²  + y²   = 4a ² + 4cx

Squaring both sides

16a²(x² + 2cx + c²  + y²) = 16a⁴ + 32a²cx + 16c²x²

∴  16a²x² + 32a²cx + 16a²c²  + 16a²y² = 16a⁴  + 32a²cx + 16c²x²

∴  16a²x² – 16c²x² + 16a²y² = 16a⁴  – 16a²c²

∴  16(a² – c²)x² + 16a²y² = 16a² ( a² – c²)

Given b² = a² – c²

∴  16b²x² + 16a²y² = 16a² b²

∴  b²x² + a²y² = a² b²

Dividing both sides by a² b²

This is the equation of locus.

Example – 43 :

Find the equation of locus of a point P so that the segment joining the points (3, 2) and (-5, 1) subtends a right angle at the point P.

Solution:

Method – I (Using Pythagoras Theorem):

Let P(x. y) be the point on the locus, A(3, 2) and B(-5, 1) be the points.

Seg AB subtends right angle at point P, hence ΔPAB is right-angled triangle

Seg AB is the hypotenuse

PA² +  PB² = AB²

∴  (x – 3)² + (y – 2)² + (x + 5)² + (y – 1)² = (3 + 5)² + (2 – 1)²

∴  x² – 6x + 9 + y² – 4y + 4 + x² + 10x + 25 + y² – 2y + 1 = 64 + 1

∴ 2 x² + 2y² + 4x  – 6y + 39 – 65 = 0

∴ 2 x² + 2y² + 4x  – 6y – 26 = 0

∴ x² + y² + 2x  – 3y – 13 = 0

Hence required equation of the locus  is  x² + y² + 2x  – 3y – 13 = 0

Method – II (Using Slopes):

Let P(x. y) be the point on the locus, A(3, 2) and B(-5, 1) be the points.

Seg AB subtends right angle at point P

Slope of AP ×  Slope of BP = -1

∴  y² – y – 2y + 2 = -(x² +5x – 3x -15)

∴  y² – 3y + 2 = – (x² + 2x -15)

∴  y² – 3y + 2 =  – x² –  2x +15

∴  y² – 3y + 2 + x² +  2x – 15 = 0

∴  x² + y² + 2x  – 3y – 13 = 0

Hence required equation of the locus  is  x² + y² + 2x  – 3y – 13 = 0

Example – 44 :

A(2, 0) and B(-2, 0) are two points. Find the equation of locus of point P, such that ∠ APB is a right angle.

Solution:

Method – I (Using Pythagoras Theorem):

Let P(x. y) be the point on the locus, A(2, 0) and B(-2, 0) be the points.

∠ APB is a right angle, hence ΔPAB is right-angled triangle

Seg AB is the hypotenuse

PA² +  PB² = AB²

∴  (x – 2)² + (y – 0)² + (x + 2)² + (y – 0)² = (2 + 2)² + (0 – 0)²

∴  x² – 4x + 4 + y² + x² + 4x + 4 + y²  = 16

∴ 2 x² + 2y²  = 8

∴  x² +  y² = 4

Hence required equation of the locus  is    x² +  y² = 4

Method – II (Using Slopes):

Let P(x. y) be the point on the locus, A(2, 0) and B(-2, 0) be the points.

Seg AB subtends right angle at point P

Slope of AP ×  Slope of BP = -1

∴  y²  = -(x²  – 4)

∴  y²  = – x²  + 4

∴   x²  +  y²  = 4

Hence required equation of the locus  is   x²  +  y²  = 4

Example – 45 :

A(5, -3) and B(-1, -5) are two points. Find the equation of locus of point P, such that seg AB subtends a right angle at point P.

Solution:

Let P(x. y) be the point on the locus, A(5, -3) and B(-1, -5) be the points.

The seg AB subtends right angle at point P, hence ΔPAB is right-angled triangle

Seg AB is the hypotenuse

PA² +  PB² = AB²

∴  (x – 5)² + (y + 3)² + (x + 1)² + (y + 5)² = (5 + 1)² + (-3 + 5)²

∴  x² – 10x + 25 + y² + 6y + 9 + x² + 2x + 1 + y² + 10y + 25  = 36  + 4

∴  2x² + 2y²  – 8x + 16 y + 60  = 40

∴  2x² + 2y²  – 8x + 16 y + 20 = 0

∴  x² + y²  – 4x + 8 y + 10 = 0

Hence required equation of the locus  is  x² + y²  – 4x + 8 y + 10 = 0

Science > Mathematics > Coordinate Geometry > Locus > Equation of Locus

The post Problems on Equation of Locus: Set – II appeared first on The Fact Factor.

In this article, we shall study the concept of locus and to find the equation of the locus.

Locus: A set of points satisfying some geometrical condition or conditions is called as the locus

For example, a circle is a locus of all the points in the plane which are equidistant from the point in the plane.

Equation of Locus: The equation of locus is an equation which is satisfied by all the points satisfying given the geometrical condition in the problem

Steps Involved in Finding Equation of Locus:

1. Assume the locus point P(x, y)
2. Write  given geometrical condition
3. Use distance, section, centroid, and other formulae as per condition
4. Substitute in geometrical condition. Simplify to get the equation of locus

Example – 01:

Find the locus of a point P such that its ordinate is 5.

Solution:

Let P(x. y) be the point on the locus

Given ordinate of P is  5 i.e. y = 5

Hence required equation of the locus of point P is y = 5.

Example – 02:

Find the locus of a point P such that its abscissa is 3.

Solution:

Let P(x. y) be the point on the locus

Given abscissa of P  is  3 i.e. x = 3

Hence required equation of the locus of point P is y = 3

Example – 03:

Find the locus of a point P such that its ordinate is equal to the abscissa.

Solution:

Let P(x. y) be the point on the locus

Given ordinate of P is equal to the abscissa

Hence required equation of the locus of point P is y = x  or x – y = 0

Example – 04:

Find the locus of a point P such that its ordinate exceeds 5 times its abscissa by 9

Solution:

Let P(x. y) be the point on the locus

Given ordinate of P exceeds 5 times its abscissa by 9

Hence required equation of the locus of point P is y = 5x + 9.

Example – 05:

Find the locus of a point P such that its abscissa exceeds 2 times its abscissa by 3

Solution:

Let P(x. y) be the point on the locus

Given abscissa of P exceeds 2 times its abscissa by 3

Hence required equation of the locus of point P is x = 2y + 3.

Example – 06:

Find the locus of a point P such that the sum of its coordinates is 15

Solution:

Let P(x. y) be the point on the locus

Given the sum of its coordinates is 15

Hence required equation of the locus of point P is x + y = 15

Example – 07:

Find the locus of a point P such that the sum of its coordinates is 10

Solution:

Let P(x. y) be the point on the locus

Given the sum of  its coordinates is 10

Hence required  inequality of the locus of point P is x + y = 10

Example – 08:

Find the locus of a point P such that the sum of its coordinates is less than 10

Solution:

Let P(x. y) be the point on the locus

Given the sum of its coordinates is less than 10

Hence required  inequality of the locus of point P is x + y < 10

Example – 09:

Find the locus of a point P such that the sum of its coordinates is greater than 5

Solution:

Let P(x. y) be the point on the locus

Given the sum of its coordinates is greater than 5

Hence required  inequality of the locus of point P is x + y > 5

Example – 10:

Find the locus of a point P such that the sum of squares of  its coordinates is 25

Solution:

Let P(x. y) be the point on the locus

Given the sum of squares of  its coordinates is 25

Hence required equation of the locus of point P is x² + y² = 25

Example – 11:

Find the locus of a point P such that the sum of square its coordinates is 9

Solution:

Let P(x. y) be the point on the locus

Given the sum of square of its coordinates is 9

Hence required  inequality of the locus of point P is x² + y² = 9

Example – 12:

Find the locus of a point P such that twice the ordinate of P exceeds thrice its abscissa by 4.

Solution:

Let P(x. y) be the point on the locus

Given twice the ordinate of P exceeds thrice its abscissa by 4.

Hence required equation of the locus of point P is 2y = 3x + 4

Example – 13:

Find the locus of a point P such that the distance of P from x-axis equal to 10 times its distance from the y-axis.

Solution:

Let P(x. y) be the point on the locus

Distance from x-axis = y

Distance from y-axis = x

Given the distance of P from x-axis equal to 10 times its distance from y-axis

Hence required equation of the locus of point P is y = 10x

Example – 14:

Find the locus of a point P such that the distance of P from x-axis equal to its distance from the y-axis.

Solution:

Let P(x. y) be the point on the locus

Distance from x-axis = y

Distance from y-axis = x

Given the distance of P from x-axis equal to its distance from y-axis

Hence required equation of the locus of point P is y = x

Example – 15:

Find the locus of a point P such that the distance of P from origin equals 5 times its distance from the point (3, -2)

Solution:

Let P(x. y) be the point on the locus, Let O(0, 0) be the origin and A(3, -2) be the point

Given OP = 5 PA

∴ OP² = 25 PA²

∴ (x – 0)²  + (y – 0)²  = 25[(x – 3)²  + (y + 2)²]

∴ x²  + y²  = 25[x² – 6x + 9  + y² + 4y + 4]

∴ x²  + y²  = 25x² – 150x + 225  + 25y² + 100y + 100

∴  25x² – 150x + 225  + 25y² + 100y + 100 – x²  – y²  = 0

∴  24x² + 24y² – 150x + 100y + 325  = 0

Hence required equation of the locus  is 24x² + 24y² – 150x + 100y + 325  = 0

Example – 16:

Find the equation of locus of a point which is equidistant from the points (2, 3) and (-4, 5)

Solution:

Let P(x. y) be the point on the locus, Let A(2, 3) and B(-4, 5) be the given points

Given PA = PB

∴ PA² = PB²

∴ (x – 2)²  + (y – 3)²  = (x + 4)²  + (y – 5)²

∴ x²  – 4x + 4 + y²  -6y + 9 = x² + 8x + 16  + y² – 10y + 25

∴ x²  – 4x + 4 + y²  -6y + 9 – x² – 8x – 16  – y² + 10y – 25 = 0

∴ – 12x   + 4y  -28 = 0

∴ 3x   – y  + 7 = 0

Hence required equation of the locus  is 3x   – y  + 7 = 0

Example – 17:

Find the equation of locus of a point which is equidistant from the points (1, 2) and (3, 4)

Solution:

Let P(x. y) be the point on the locus, Let A(1, 2) and B(3, 4) be the given points

Given PA = PB

∴ PA² = PB²

∴ (x – 1)²  + (y – 2)²  = (x – 3)²  + (y – 4)²

∴ x²  – 2x + 1 + y²  – 4y + 4 = x² – 6x + 9  + y² – 8y + 16

∴ x²  – 2x + 1 + y²  -4y + 4 – x² + 6x – 9  – y² + 8y – 16 = 0

∴ – 2x + 1 -4y + 4 + 6x – 9 + 8y – 16 = 0

∴ 4x   + 4y  – 20 = 0

∴ x   + y  – 5 = 0

Hence required equation of the locus  is x   + y  – 5 = 0

Example – 18:

Find the equation of locus of a point which is equidistant from the points (2, 3) and (5, 7)

Solution:

Let P(x. y) be the point on the locus, Let A(2, 3) and B(5, 7) be the given points

Given PA = PB

∴ PA² = PB²

∴ (x – 2)²  + (y – 3)²  = (x – 5)²  + (y – 7)²

∴ x²  – 4x + 4 + y²  – 6y + 9 = x² – 10x + 25  + y² – 14y + 49

∴ x²  – 4x + 4 + y²  – 6y + 9 – x² + 10x – 25  – y² + 14y – 49 = 0

∴ 6x  +   8y  – 61 = 0

Hence required equation of the locus  is 6x  +   8y  – 61 = 0

Example – 19:

A(2, 3), B(-2, 5) are given points. Find the equation of locus of point P, such that PA = 2PB.

Solution:

Let P(x. y) be the point on the locus, Given A(2, 3) and B(-2, 5) be the given points

Given PA = 2 PB

∴ PA² = 4 PB²

∴ (x – 2)²  + (y – 3)²  = 4[(x + 2)²  + (y – 5)²]

∴ x²  – 4x + 4 + y²  – 6y + 9 = 4[ x² + 4x + 4  + y² – 10y + 25]

∴ x²  – 4x + 4 + y²  – 6y + 9 = 4 x² + 16x + 16  + 4y² – 40y + 100

∴  4 x² + 16x + 16  + 4y² – 40y + 100 – x²  + 4x – 4 – y²  + 6y – 9 = 0

∴  3 x² + 3y² + 20x – 34y + 103 = 0

Hence required equation of the locus  is 3 x² + 3y² + 20x – 34y + 103 = 0

Example – 20:

A(2, 3), B(-2, 1) are given points. Find the equation of locus of point P, such that AP² = 3 BP².

Solution:

Let P(x. y) be the point on the locus, Given A(2, 3) and B(-2, 1) be the given points

Given AP² = 3 BP²

∴ (x – 2)²  + (y – 3)²  = 3[(x + 2)²  + (y – 1)²]

∴ x²  – 4x + 4 + y²  – 6y + 9 = 3[ x² + 4x + 4  + y² – 2y + 1]

∴ x²  – 4x + 4 + y²  – 6y + 9 = 3x² + 12x + 12  + 3y² – 6y + 3

∴ 3x² + 12x + 12  + 3y² – 6y + 3 –  x²  + 4x – 4 – y²  + 6y – 9 = 0

∴ 2x² + 2y² + 16x + 2 = 0

∴ x² + y² + 8x + 1 = 0

Hence required equation of the locus  is x² + y² + 8x + 1 = 0

Example – 21:

A(3, 1), B(4, -5) are given points. Find the equation of locus of point P, such that AP²+ BP² = 50.

Solution:

Let P(x. y) be the point on the locus, Given A(3, 1) and B(4, -5) be the given points

Given AP²+ BP² = 50.

∴ (x – 3)²  + (y – 1)²  +  (x – 4)²  + (y + 5)² = 50

∴  x²  – 6x + 9 + y²  – 2y + 1 +  x² – 8x + 16  + y² + 10y + 25 = 50

∴  2x² + 2y² – 14x  + 8y + 51= 50

∴  2x² + 2y² – 14x  + 8y + 1 = 0

Hence required equation of the locus  is 2x² + 2y² – 14x  + 8y + 1 = 0

Example – 22:

A(-3, 2), B(1, -4) are given points. Find the equation of locus of point P, such that 3PA = 2 PB

Solution:

Let P(x. y) be the point on the locus, Given A(-3, 2) and B(1, -4) be the given points

Given 3 PA = 2 PB

∴ 9 PA²= 4 PB²

∴9[ (x + 3)²  + (y – 2)²]  =  4[(x – 1)²  + (y + 4)²]

∴ 9( x²  + 6x + 9 + y²  – 4y + 4) = 4(x² – 2x + 1  + y² + 8y + 16)

∴ 9 x²  + 54x + 81 + 9y²  – 36y +36 = 4x² – 8x + 4  + 4y² + 32y + 64

∴ 9 x²  + 54x + 81 + 9y²  – 36y +36 – 4x² + 8x – 4  – 4y² – 32y – 64 = 0

∴ 5x²  + 5y²  + 62x  – 68y +49 = 0

Hence required equation of the locus  is 5x²  + 5y²  + 62x  – 68y +49 = 0

Example – 23:

Find the equation perpendicular bisector of segment joining points (-5, 2) and (-1, 5)

Solution:

Let P(x. y) be the point on the locus, Let A(-5, 2) and B(-1, 5) be the given points

Perpendicular bisector of a line segment is locus of all the points

in the plane whci is equidistant from the endpoints of the segment

Hence PA = PB

∴ PA² = PB²

∴ (x + 5)²  + (y – 2)²  = (x + 1)²  + (y – 5)²

∴ x²  + 10x + 25 + y²  -4y + 4 = x² + 2x + 1 + y² – 10y + 25

∴ x²  + 10x + 25 + y²  -4y + 4 – x² – 2x – 1 – y² + 10y – 25 = 0

∴ 8x  + 6y + 3 = 0

Hence required equation of the locus (perpendicular bisector) is 3x   – y  + 7 = 0

Example – 24:

A(4, 5), B(-2, 7) are given points. Find the equation of locus of point P, such that 2PA = PB

Solution:

Let P(x. y) be the point on the locus, Given A(4, 5) and B(-2, 7) be the given points

Given 2PA = PB

∴ 4 PA²= PB²

∴4[ (x -4 )²  + (y – 5)²]  =  (x + 2)²  + (y – 7)²

∴ 4( x²  – 8x + 16 + y²  – 10y + 25) = x² + 4x + 4  + y² – 14y + 49

∴ 4x²  – 32x +64 + 4y²  – 40y + 100 = x² + 4x + 4  + y² – 14y + 49

∴ 4x²  – 32x +64 + 4y²  – 40y + 100 – x² – 4x – 4  – y² + 14y – 49 = 0

∴ 3x²  + 3y²  – 36x  – 26 y +111 = 0

Hence required equation of the locus  is 3x²  + 3y²  – 36x  – 26 y + 111 = 0

Example – 25:

Find the equation of locus of points whose distance from the point (2, -3) is twice its distance from (1, 2).

Solution:

Let P(x. y) be the point on the locus, Given A(2, -3) and B(1, 2) be the given points

Given PA = 2 PB

∴ PA²= 4 PB²

∴ (x – 2 )²  + (y + 3)²  = 4[ (x – 1)²  + (y – 2)²]

∴  x²  – 4x + 4 + y²  + 6y + 9 = 4[x² – 2x + 1  + y² – 4y + 4]

∴  x²  – 4x + 4 + y²  + 6y + 9 = 4x² – 8x + 4  + 4y² – 16y + 16

∴  4x² – 8x + 4  + 4y² – 16y + 16 –  x² + 4x – 4 – y²  – 6y – 9 = 0

∴ 3x²  + 3y²  – 4x  – 22 y + 7 = 0

Hence required equation of the locus  is 3x²  + 3y²  – 4x  – 22 y + 7 = 0

Example – 26:

Find the equation of locus of a point such that, the difference of square of its distances from the points (5, 0) and (2, 3) is 10.

Solution:

Let P(x. y) be the point on the locus, Given A(5, 0) and B(2, 3) be the given points

Given  PA² – PB² = 10

∴ [(x – 5 )²  + (y – 0)² ] –  [(x – 2)²  + (y – 3)²] = 10

∴ ( x²  – 10x + 25 + y²) – (x² – 4x + 4  + y² – 6y + 9) = 10

∴ x²  – 10x + 25 + y² – x² + 4x – 4  – y² + 6y – 9 – 10 = 0

∴  – 6x + 6y + 2 = 0

∴  3x – 3y – 1 = 0

Hence required equation of the locus  is 3x – 3y – 1 = 0

Example – 27:

Find the equation of locus of a point such that, the sum of the square of its distances from the points (2, -3) and (-1, -2) is 15.

Solution:

Let P(x. y) be the point on the locus, Given A(2, -3) and B(-1, -2) be the given points

Given  PA² + PB² = 15

∴ [(x – 2 )²  + (y + 3)² ] +  [(x + 1)²  + (y + 2)²] = 15

∴ x²  – 4x + 4 + y² + 6y + 9 + x² + 2x + 1  + y² + 4y +4 – 15 = 0

∴ 2x² + 2y²  – 2x +10y +  3 = 0

Hence required equation of the locus  is 2x² + 2y²  – 2x +10y +  3 = 0

Example – 28:

A(5, -6), B(-1, 2) and C(4, -3) are three given points. Find the equation of locus of points P such that PA² + PC² = AB².

Solution:

Let P(x. y) be the point on the locus, Given A(5, -6), B(-1, 2) and C(4, -3) be the given points

Given  PA² + PC² = AB²

∴ [(x – 5 )²  + (y + 6)² ] +  [(x – 4)²  + (y + 3)²] = (-1 – 5)² + (2 + 6)²

∴ x²  – 10x + 25 + y² + 12y + 36 + x² – 8x + 16  + y² + 6y +9 = 36 + 64

∴ 2x²  + 2 y² – 18x + 18 y – 14 = 0

∴ x²  + y² – 9x + 9y – 7 = 0

Hence required equation of the locus  is x²  + y² – 9x + 9y – 7 = 0

Example – 29:

A(5, -6), B(-1, 2) and C(4, -3) are three given points. Find the equation of locus of points P such that PA² – PB² = 12.

Solution:

Let P(x. y) be the point on the locus, Given A(5, -6), B(-1, 2) and C(4, -3) be the given points

Given  PA² – PB² = 12

∴ [(x – 5 )²  + (y + 6)² ] –  [(x + 1)²  + (y – 2)²] =12

∴ (x²  – 10x + 25 + y² + 12y + 36) – (x² + 2x + 1  + y² – 4y + 4) = 12

∴ x²  – 10x + 25 + y² + 12y + 36 – x² – 2x –  1  – y² + 4y – 4 – 12 = 0

∴ – 12x  + 16y  + 44 = 0

∴ 3x  – 4y  – 11 = 0

Hence required equation of the locus  is 3x  – 4y  – 11 = 0

Example – 30 :

Find the equation of locus of a point such that its distance from the origin is three times its distance from the x-axis.

Solution:

Let P(x. y) be the point on the locus, O(0. 0) be the origin

Distance of point from x-axis = y

Given: OP = 3y

∴ OP² = 9y²

∴ [(x – 0 )²  + (y – 0)² ] = 9y²

∴ x ²  + y ²  = 9y²

∴ x ²  –  8y ²  = 0

Hence required equation of the locus  is x ²  –  8y ²  = 0

Science > Mathematics > Coordinate Geometry > Locus > Equation of Locus

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