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