# Point on Locus

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#### 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 – 1:

• 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 – 2:

• 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 – 3:

• 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 – 5:

• 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 – 6:

• 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 – 7:

• 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 – 8:

• 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 – 9:

• 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)

#### 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 – 10:

• 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 – 11:

• 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 – 12:

• 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 – 13:

• 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 – 14:

• 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 – 15:

• 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 – 16:

• Find the point of intersection of the locus x2 + y2 – 4x – 6y -12 = 0 with x- axis and find the length of intercept made by the curve on 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)= 64

∴  AB = 8

Thus point of intersection of 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 – 17:

• Find the point of intersection of the locus 16x2 + 25y2 = 400 with x- axis and find the length of intercept made by the curve on 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)= 100

∴  AB = 1

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 – 18:

• 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)

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### One Comment

1. Akib khan

That a great