Students must practice these TS Intermediate Maths 1B Solutions Chapter 1 Locus Ex 1(a) to find a better approach to solving the problems.

## TS Inter 1st Year Maths 1B Locus Solutions Exercise 1(a)

Question 1.

Find the equation of locus of a point which is at a distance 5 from A (4, – 3). (V.S.A.Q)

Answer:

Given A = (4, – 3) and suppose P (x, y) is any other point on the locus.

Then given PA = 5

⇒ PA^{2} = 25

⇒ (x – 4)^{2} + (y + 3)^{2} = 25

⇒ x^{2} + y^{2} – 8x + 6y + 25 – 25 = 0

⇒ x^{2} + y^{2} – 8x + 6y = 0 ……………… (1)

(If there exists another point Q(x_{1}, y_{1}) such that QA^{2} = (x_{1} – 4)^{2} + (y_{1} + 3)^{2}

Let Q(x_{1}, y_{1} satisfy (1) then

x_{1}^{2} + y_{1}^{2} – 8x_{1} + 6y_{1} = 0

and QA^{2} = x_{1}^{2} + y_{1}^{2} – 8x_{1} + 6y_{1} + 25

= 0 + 25 = 25

⇒ QA = 5

∴ Q (x_{1}, y_{1}) satisfy the geometric condition (1)

∴ Required equation of locus is x^{2} + y^{2} – 8x + 6y = 0

Note : Second part need not follow the problem from the definition of locus.

Question 2.

Find the equation of locus of a point which is equidistant from the points A (-3, 2) and B (0, 4). (V.S.A.Q)

Answer:

Let P (x, y) be any point on the locus. Then from the given geometric condition PA = PB

∴ PA^{2} = PB^{2}

⇒ (x + 3)^{2} + (y – 2)^{2} = (x – 0)^{2} + (y – 4)^{2}

⇒ x^{2} + y^{2} + 6x + 9 – 4y + 4 = x^{2} + y^{2} – 8y + 16

⇒ 6x + 4y – 3 = 0

⇒ 6x + 4y = 3 is the equation of the locus.

Question 3.

Find the equation of locus of a point P such that the distance of P from the origin is twice the distance of P from A (1, 2).

(V.S.A.Q) (March 2012)

Answer:

Given O (0,0) and A (1,2) are the two points, P (x, y) be any point on the locus. From the given condition OP = 2PA

⇒ OP^{2} = 4 PA^{2}

⇒ x^{2} + y^{2} = 4[(x – 1)^{2} + (y – 2)^{2}]

⇒ x^{2} + y^{2} = 4 [(x^{2} + y^{2} – 2x – 4y + 1 + 4]

⇒ 4 [x^{2} + y^{2} – 2x – 4y + 5]

⇒ 4x^{2} + 4y^{2} – 8x – 16y + 20

∴ Equation to the locus of P is

3x^{2} + 3y^{2} – 8x – 16y + 20 = 0

Question 4.

Find the equation of locus of a point which is equidistant from the coordinate axes. (V.S.A.Q)

Answer:

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

The distance from P to X – axis is ’y and that of the distance to Y – axis is ‘x’.

Given y = x ⇒ y^{2} = x^{2}

locus of P (x, y) is x^{2} – y^{2} = 0

Question 5.

Find the equation of locus of a point equidistant from A (2, 0) and the Y – axis. (V.S.A.Q)

Answer:

A (2, 0) is the given point and

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

The distance from P to Y – axis is PB = x

Given PA = PB

⇒ PA^{2} = PB^{2}

⇒ (x – 2)^{2} + y^{2} = x^{2}

⇒ x^{2} – 4x + 4 + y^{2} = x^{2}

⇒ y^{2} – 4x + 4 = 0

∴ Locus of P (x, y) is y – 4x + 4 = 0

Question 6.

Find the equation of locus of a point P the square of whose distance from the origin is 4 times its y – coordinate. (V.S.A.Q)

Answer:

Let P (x, y) be any point on the locus. Its distance from origin is OP

Given that OP^{2} = 4y

⇒ x^{2} + y^{2} = 4y

⇒ x^{2} + y^{2} – 4y = 0

∴ Equation to the locus of P is

x^{2} + y^{2} – 4y = 0

Question 7.

Find the equation of locus of a point P Such that PA^{2} + PB^{2} = 2c^{2} where A = (a, 0), B= (- a, 0) and 0 < |a| < |c| (V.S.A.Q)

Answer:

Let P (x, y) be any point on the locus. Given A = (a, 0) and B = (-a, 0) are two points. Given condition is PA^{2} + PB^{2} = 2c^{2}

⇒ (x – a)^{2} + y^{2} + (x + a)^{2} + y^{2} = 2c^{2}

⇒ x^{2} – 2ax + a^{2} + y^{2} + x^{2} + 2ax + a^{2} + y^{2} = 2c^{2}

⇒ 2x^{2} + 2y^{2} = 2c^{2} – 2a^{2}

∴ x^{2} + y^{2} = c^{2} – a^{2} is the equation of locus of P.

II.

Question 1.

Find the equation of locus of P, if the line segment joining (2, 3) and (-1,5) subtends a right angle at P. (May ’12, March ’13, ’05) (S.A.Q)

Answer:

A = (2, 3), B = (-1, 5) are the given points. P (x, y ) is any point on the locus.

Given condition is ∠APB = 90°

Using pythagorous theorem, AP^{2} + PB^{2} = AB^{2}

⇒ (x – 2)^{2} + (y – 3)^{2} + (x + 1)^{2} + (y – 5)^{2} = ( 2 + 1)^{2} + ( 3 – 5)^{2}

⇒ x2^{2} – 4x + 4 + y^{2} – 6y + 9 + x^{2} + 2x + 1 + y^{2} – 10y + 25 = 9 + 4

⇒ 2x^{2} + 2y^{2} – 2x – 16y + 26 = 0

∴ Locus of P is x^{2} + y^{2} – x – 8y + 13 = 0

(x, y) ≠ (2, 3) and (x, y) ≠ (-1,5)

Question 2.

The ends of the hypotenuse of a right angled triangle are (0, 6 ) and ( 6, 0 ). Find the equation of locus of its third vertex. (S.A.Q)

Answer:

The ends of the hypotenuse are given as A (0, 6) and B (6, 0)

Let P (x, y) be the third vertex.

Given condition is ∠APB = 90°

∴ By pythagorous theorem

⇒ AP^{2} + PB^{2} = AB^{2}

⇒(x – 0)^{2} + (y – 6)^{2} + (x – 6)^{2} + (y – 0)^{2} = (0 – 6)^{2} + (6 – 0)^{2}

⇒ 2x^{2} + 2y^{2} – 12y – 12x + 36 + 36 = 36 + 36

⇒ 2x^{2} + 2y^{2} – 12x – 12y = 0

∴ Locus of P (x, y) is x^{2} + y^{2} – 6x – 6y = 0

(x, y) ≠ (0, 6) and (x, y) ≠ (6, 0)

Question 3.

Find the equation of locus of a point the difference of whose distances from (- 5, 0) and (5, 0) is 8. (May 2011,’06, March 2001) (S.A.Q)

Answer:

A (- 5, 0) and B (5, 0) are the given points.

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

From the given condition |PA – PB| = 8 …… (1)

Consider

PA^{2} – PB^{2} = [(x + 5)^{2} + (y – 0)^{2}] – [(x – 5)^{2} + (y – 0)^{2}]

= (x^{2} + 10x + 25 + y^{2}) – (x^{2} – 10x + 25 + y^{2}) = – 20x

∴ (PA + PB) (PA – PB) = 20x

⇒ (PA + PB) (8) = 20x

⇒ PA + PB = \(\frac{5}{2}\) x .

Adding (1) and (2)

2PA = 8 + \(\frac{5 x}{2}\)x

⇒ 4PA = 16 – 5x

⇒ 16 PA = (16 – 5x)^{2}

⇒ 16 [(x + 5)^{2} + y^{2}] = (16 + 5x)^{2}

⇒ 16 [x^{2} + y^{2} + 10x + 25] = (16 + 5x)^{2}

⇒ 16x^{2} + 16y^{2} + 160x + 400 = 256 + 160x + 25^{2}

⇒ 9x^{2} + 16y^{2} + 144 = 0

⇒ 9x^{2} – 16y^{2} = 144

⇒ \(\frac{9 x^2}{144}-\frac{16 y^2}{144}\) = 1

⇒ \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1

∴ Equation of locus of P(x, y) is \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1

Question 4.

Find the equation of locus of P, if

A = (4, 0), B = (- 4, 0) and |PA – PB| = 4. (S.A.Q) (May’ 2007)

Answer:

Given that A = (4, 0) and B = (- 4, 0) are two points and let P (x ,y) be any point on the locus.

The given condition is |PA – PB| = 4 ………………. (1)

Consider

PA^{2} – PB^{2} = [(x – 4)^{2} + y^{2}] – [(x + 4)^{2} + y^{2}]

= (x^{2} – 8x + 16 + y^{2}) – (x^{2} + y^{2} + 8x + 16)

= – 16x

(PA + PB) (PA – PB) = – 16x

⇒ (PA + PB) (4) = -16x

⇒ PA + PB = – 4x ……………….. (2)

Adding (1) and (2)

2PA = 4 – 4x

⇒ PA = 2 – 2x

⇒ PA^{2} = (2 – 2x)^{2}

⇒ (x – 4)^{2} + y^{2} = 4 – 8x + 4x^{2}

⇒ x^{2} + y^{2} – 8x + 16 = 4x^{2} – 8x + 4

⇒ 3x^{2} – y^{2} = 12

⇒ \(\frac{x^2}{4}-\frac{y^2}{12}\) = 1

∴ Equation to the locus of P is \(\frac{x^2}{4}-\frac{y^2}{12}\) = 1

Question 5.

Find the equation of locus of a point, the sum of whose distances from (0, 2) and (0, – 2) is 6 . (S.A.Q)

Answer:

Let A = (0, 2) and B = (0, – 2) are the two given points.

Let P (x, y) b.e any point on the locus.

From the given condition, PA + PB = 6 …………….. (1)

Consider

PA^{2} – PB^{2} = [(x – 0)^{2} + (y – 2)^{2}] – [(x – 0)^{2} + (y + 2)^{2}]

= x^{2} + y^{2} – 4y + 4 – x^{2} – y^{2} – 4y – 4 = – 8y

∴ (PA + PB) (PA – PB) = – 8y

⇒ 6 (PA – PB) = – 8y

⇒ (PA – PB) = \(\frac{-8 y}{6}=\frac{-4 y}{3}\) ………………… (2)

Adding (1) and (2)

⇒ 9x^{2} + 9y^{2} + 36 = 81 + 4y^{2}

⇒ 9x^{2} + 5y^{2} = 45

⇒ \(\frac{x^2}{5}+\frac{y^2}{9}\) = 1

∴ Equation to the locus of P is \(\frac{x^2}{5}+\frac{y^2}{9}\) = 1

Question 6.

Find the equation of the locus of P, if A = ( 2, 3 ), B = ( 2, -3 ) and PA + PB = 8. (SA.Q)

Answer:

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

Given condition is PA + PB = 8 (1)

PA^{2} – PB^{2} = [(x – 2)^{2} + (y – 3)^{2}] – [ (x – 2)^{2} + ( y + 3)^{2}]

= (x – 2)^{2} + (y – 3)^{2} – (x – 2)^{2} – (y + 3)^{2} = – 12y

(y – 3)^{2} – (y + 3)^{2} = – 12y

∴ PA^{2} – PB^{2} = – 12y

(PA + PB) (PA – PB) = -12y

⇒ 8 (PA – PB) = – 12y

⇒ PA – PB = \(\frac{-12 y}{8}=\frac{-3 y}{2}\) ……………….. (2)

Adding (1) and (2)

2PA = 8 – \(\frac{3 \mathrm{y}}{2}\) = \(\frac{16-3 y}{2}\)

⇒ 4PA = 16 – 3y

⇒ 16PA^{2} = (16 – 3y)^{2}

⇒ 16 [(x – 2)^{2} + (y – 3)^{2}] = (16 – 3y)^{2}

⇒ 16 [x^{2} + y^{2} – 4x – 6y + 13] = 256 – 96y + 9y^{2}

⇒ 16x^{2} + 7y^{2} – 64x – 48 = 0

∴ Equation to the locus of P is

16x^{2} + 7y^{2} – 64x – 48 = 0

(or) 16(x^{2} – 4x) + 7y^{2} = 48

⇒ 16 (x^{2} – 4x + 4) + 7y^{2} = 112

⇒ 16 (x – 2)^{2} + 7y^{2} = 112.

⇒ \(\frac{(x-2)^2}{7}+\frac{y^2}{16}\)

Question 7.

A (5, 3) and B (3, – 2) are two fixed points. Find the equation to the locus of P, so that the area of triangle PAB is 9. (S.A.Q) (March 2006)

Answer:

A (5, 3 ), B (3, -2 ) are the given points.

Let P (x, y ) be any point on the locus. Given condition is that the area of ∆ PAB = 9

⇒ \(\frac{1}{2}\) |[x_{1} (y_{2} – y_{3}) + x_{2} (y_{3} – y_{1}) + x_{3} (y_{1} – y_{2})[|

= \(\frac{1}{2}\) |[5(- 2 – y) + 3 (y – 3) + x (3 + 2)]| = 9

⇒ 5x – 2y – 19 = ±18

⇒ 5x – 2y – 19 = 18 (or) 5x – 2y – 19 = – 18

⇒ 5x – 2y – 37 = 0 (or) 5x – 2y – 1= 0

∴ Locus of P is (5x – 2y – 37) (5x – 2y – 1) = 0

Question 8.

Find the equation of locus of a point, which forms a triangle of area 2 with the points A (1, 1 ) and B ( – 2, 3 ). (S.A.Q)

Answer:

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

Given condition is ∆ PAB = 2

⇒ ∴ \(\frac{1}{2}\) |1(3 – y) – 2(y – 1) + x(1 – 3)| = 2

⇒ |3 – y – 2y + 2 – 2x| = 4

⇒ – 2x – 3y + 5 = ±4

⇒ – 2x – 3y + 5 = 4 (or) – 2x – 3y + 5 = – 4

⇒ 2x + 3y – 1 = 0 (or) 2x + 3y – 9 = 0

∴ Locus of P (x, y) is (2x + 3y – 1) (2x + 3y – 9) = 0

Question 9.

If the distance from P to the points (2, 3) and ( 2, – 3 ) are in the ratio 2 : 3 then find the equation of locus of P. (S.A.Q) (May 2014, March 2014)

Answer:

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

Given points are A (2, 3) and B (2, -3) and given condition is PA : PB = 2: 3

⇒ 3PA = 2PB ⇒ 9PA^{2} = 4PB^{2}

⇒ 9 [(x – 2)^{2} + (y – 3)^{2}] = 4 [(x – 2)^{2} + (y + 3)^{2}]

⇒ 9 [(x^{2} – 4x + 4 + y^{2} – 6y + 9)] = 4 [x^{2} – 4x + 4 + y^{2} + 6y + 9]

⇒ 5x^{2} + 5y^{2} – 20x – 78y + 65 = 0

∴ Equation to the locus of P is

5x^{2} + 5y^{2} – 20x – 78y + 65 = 0

Question 10.

A (1, 2 ), B ( 2, – 3) and C (- 2, 3) are three points. A point P moves such that PA^{2} + PB^{2} = 2PC^{2}. Show that the equation to the locus of P is 7x – 7y + 4 = 0 (S.A.Q) (May 2007)

Answer:

Let P (x, y) be any point on the locus. Given points are

A = (1, 2) ; B = (2, – 3) and C = (- 2, 3)

Given condition is PA^{2} + PB^{2} = 2PC^{2}

⇒[(x – 1)^{2} + (y – 2)^{2}] + [(x – 2)^{2} + (y + 3)^{2}]

= 2 [(x + 2)^{2} + (y – 3)^{2}]

⇒ 2x^{2} + 2y^{2} – 6x + 2y + 18

= 2x^{2} + 2y^{2} + 8x – 12y + 26

⇒ 14x – 14y + 8 = 0

⇒ 7x – 7y + 4 = 0

∴ Equation to the locus of P is

7x – 7y + 4 = 0