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Science > Mathematics > Locus > You are Here |

#### To Check Whether the Point Lies on the Locus:

#### Algorithm:

- Write the equation of given locus
- To check whether the point (x
_{1}, y_{1}) lies on the locus, substitute x = x_{1}and y = y_{1}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 2x
^{2}+ 2y^{2}– 5x + 11y – 3 = 0. **Solution:**- To check point (4, -1)

The equation of the locus is 2x^{2} + 2y^{2} – 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 2x^{2} + 2y^{2} – 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 x
^{2}+ y^{2}+ 6x – 6y – 47 = 0. **Solution:**- To check point (5, 2)

The equation of the locus is x^{2} + y^{2} + 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 x^{2} + y^{2} + 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 x^{2} + y^{2} + 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(at
^{2}, 2at) lies on the locus y^{2}= 4ax **Solution:**

The equation of the locus is y^{2} = 4ax

Substituting x = at^{2}and y = 2at, in L.H.S. of equation of locus

L.H.S. = y^{2} = (2at)^{2} = 4a^{2}t^{2 }………. (1)

∴ R.H.S. = 4ax = 4a (at^{2}) = 4a^{2}t^{2 }………. (2)

From equations (1) and (2) we have

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

Hence point P(at^{2}, 2at) lies on the locus y^{2} = 4ax

#### Example – 5:

- Show that the point Q(a cosθ, b sinθ) lies on the locus x
^{2}/a^{2}+ y^{2}/b^{2}= 1 **Solution:**

The equation of the locus is x^{2}/a^{2} + y^{2}/b^{2} = 1

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

L.H.S. = (a cosθ)^{2}/a^{2} + (b sinθ)^{2}/b^{2}

L.H.S. = a^{2} cos^{2}θ/a^{2} + b^{2} sin^{2}θ/b^{2} = cos^{2}θ + sin^{2}θ = 1 = R.H.S.^{ }..

Hence point Q(a cosθ, b sinθ) lies on the locus x^{2}/a^{2} + y^{2}/b^{2} = 1

#### Example – 6:

- Show that the point R(a secθ, b tanθ) lies on the locus x
^{2}/a^{2}– y^{2}/b^{2}= 1 **Solution:**

The equation of the locus is x^{2}/a^{2} – y^{2}/b^{2} = 1

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

L.H.S. = (a secsθ)^{2}/a^{2} – (b tanθ)^{2}/b^{2}

L.H.S. = a^{2} sec^{2}θ/a^{2} – b^{2} tan^{2}θ/b^{2} = sec^{2}θ – tan^{2}θ = 1 = R.H.S.^{ }..

Hence point R(a secθ, b tanθ) lies on the locus x^{2}/a^{2} – y^{2}/b^{2} = 1

#### Example – 7:

- Show that the point S(a cosθ, a sinθ) lies on the locus x
^{2}+ y^{2}= a^{2} **Solution:**

The equation of the locus is x^{2} + y^{2} = a^{2}

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. = a^{2} cos^{2}θ + a^{2} sin^{2}= a^{2}(cos^{2}θ + sin^{2}θ) = a^{2}(1) = a^{2} =R.H.S.^{ }..

Hence point S(a cosθ, a sinθ) lies on the locus x^{2} + y^{2} = a^{2}

#### Example – 8:

- Show that the point T(5 cosθ, 5 sinθ) lies on the locus x
^{2}+ y^{2}= 25 **Solution:**

The equation of the locus is x^{2} + y^{2} = 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 cos^{2}θ + 25 sin^{2}= 25(cos^{2}θ + sin^{2}θ) = 25(1) = 25 =R.H.S.^{ }..

Hence point T(5 cosθ, 5 sinθ) lies on the locus x^{2} + y^{2} = 25

**Ans:** k = -9 and a = – 4/3

#### Example – 9:

- show that for all values of r, the point (x
_{1}+ r cosθ, y_{1}+ r sinθ) always lies on locus y – y_{1}= tanθ (x – x_{1}) **Solution:**

The equation of the locus is y – y_{1} = tanθ (x – x_{1})

Point (x_{1} + r cosθ, y_{1} + r sinθ) lies on it

Substituting x = x_{1} + r cosθ and y_{1} + r sinθ

L.H.S. = (y_{1} + r sinθ) – y_{1} = r sinθ …….. (1)

R.H.S. = tanθ ((x_{1} + r cosθ) – x_{1}) = 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 (x_{1} + r cosθ, y_{1} + r sinθ) lies on the locus y – y_{1} = tanθ (x – x_{1})

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

#### Algorithm:

- Write the equation of given locus
- Let given point be (x
_{1}, y_{1}) lies on the locus, substitute x = x_{1}and y = y_{1}in the equation of locus. - solve an equation to find the value of constant.

#### Example – 10:

- Point (-6, 3) lies on the locus x
^{2}= 4ay. Find the value of ‘a’. **Solution:**

The equation of the locus is x^{2} = 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 y
^{2}= ax. Find the value of ‘a’ and ‘b’. **Solution:**

The equation of the locus is y^{2} = 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 y^{2} = ax

b^{2} = (- 4)(- 16) = 64

∴ b = ± 8

**Ans: ** a = -4 and b = ± 8

#### Example – 13:

- Point (-8, 6) lies on locus x
^{2}/4 + y^{2}/3 = k . Find the value of ‘k’. **Solution:**

The equation of the locus is x^{2}/4 + y^{2}/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 x
^{2 }– 7x + ky = 0. Find the values of ‘k’ and ‘a’. **Solution:**

The equation of the locus is x^{2 }– 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 x
^{2}+ y^{2}– 2x – 14y -15 = 0 with a) x- axis and b) y-axis **Solution:**

Equation of the locus is x^{2} + y^{2} – 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

a^{2} + 0^{2} – 2a – 14(0) -15 = 0

∴ a^{2} – 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

0^{2} + b^{2} – 2(0) – 14(b) -15 = 0

∴ b^{2} – 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 x
^{2}+ y^{2}– 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 x^{2} + y^{2} – 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

a^{2} + 0^{2} – 4a – 6(0) -12 = 0

∴ a^{2} – 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

AB^{2} = (x_{1} – x_{2})^{2} + (y_{1} – y_{2})^{2}

∴ AB^{2} = (6 + 2)^{2} + (0 – 0)^{2 }= 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 16x
^{2}+ 25y^{2}= 400 with x- axis and find the length of intercept made by the curve on x-axis. **Solution:**

Equation of the locus is 16x^{2} + 25y^{2} = 400

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

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

16a^{2} + 25(0)^{2} = 400

∴ 16a^{2} = 400

∴ a^{2} = 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

AB^{2} = (x_{1} – x_{2})^{2} + (y_{1} – y_{2})^{2}

∴ AB^{2} = (5 + 5)^{2} + (0 – 0)^{2 }= 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 3x
^{2}– 5xy + 6y^{2}– 54 = 0. **Solution:**

Equation of the locus is 3x^{2} – 5xy + 6y^{2} – 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

∴ 6b^{2} = 54

∴ b^{2} = 9

∴ b =± 3

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

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