Science > Mathematics > Coordinate Geometry > Locus > Point on Locus
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:
- Write the equation of given locus
- 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.
- 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:
- Write the equation of given locus
- Let given point be (x1, y1) lies on the locus, substitute x = x1 and y = y1 in the equation of locus.
- 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:
- 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.
- 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)