TS Inter 1st Year Maths 1B Solutions Chapter 1 Locus Ex 1(a)

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
⇒ PA² = 25
⇒ (x – 4)² + (y + 3)² = 25
⇒ x² + y² – 8x + 6y + 25 – 25 = 0
⇒ x² + y² – 8x + 6y = 0 ……………… (1)
(If there exists another point Q(x₁, y₁) such that QA² = (x₁ – 4)² + (y₁ + 3)²
Let Q(x₁, y₁ satisfy (1) then
x₁² + y₁² – 8x₁ + 6y₁ = 0
and QA² = x₁² + y₁² – 8x₁ + 6y₁ + 25
= 0 + 25 = 25
⇒ QA = 5
∴ Q (x₁, y₁) satisfy the geometric condition (1)
∴ Required equation of locus is x² + y² – 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

∴ PA² = PB²
⇒ (x + 3)² + (y – 2)² = (x – 0)² + (y – 4)²
⇒ x² + y² + 6x + 9 – 4y + 4 = x² + y² – 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

⇒ OP² = 4 PA²
⇒ x² + y² = 4[(x – 1)² + (y – 2)²]
⇒ x² + y² = 4 [(x² + y² – 2x – 4y + 1 + 4]
⇒ 4 [x² + y² – 2x – 4y + 5]
⇒ 4x² + 4y² – 8x – 16y + 20
∴ Equation to the locus of P is
3x² + 3y² – 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 ⇒ y² = x²
locus of P (x, y) is x² – y² = 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
⇒ PA² = PB²
⇒ (x – 2)² + y² = x²
⇒ x² – 4x + 4 + y² = x²
⇒ y² – 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 OP² = 4y
⇒ x² + y² = 4y
⇒ x² + y² – 4y = 0
∴ Equation to the locus of P is
x² + y² – 4y = 0

Question 7.
Find the equation of locus of a point P Such that PA² + PB² = 2c² 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 PA² + PB² = 2c²
⇒ (x – a)² + y² + (x + a)² + y² = 2c²
⇒ x² – 2ax + a² + y² + x² + 2ax + a² + y² = 2c²
⇒ 2x² + 2y² = 2c² – 2a²
∴ x² + y² = c² – a² 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, AP² + PB² = AB²
⇒ (x – 2)² + (y – 3)² + (x + 1)² + (y – 5)² = ( 2 + 1)² + ( 3 – 5)²
⇒ x2² – 4x + 4 + y² – 6y + 9 + x² + 2x + 1 + y² – 10y + 25 = 9 + 4
⇒ 2x² + 2y² – 2x – 16y + 26 = 0
∴ Locus of P is x² + y² – 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
⇒ AP² + PB² = AB²
⇒(x – 0)² + (y – 6)² + (x – 6)² + (y – 0)² = (0 – 6)² + (6 – 0)²
⇒ 2x² + 2y² – 12y – 12x + 36 + 36 = 36 + 36
⇒ 2x² + 2y² – 12x – 12y = 0
∴ Locus of P (x, y) is x² + y² – 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
PA² – PB² = [(x + 5)² + (y – 0)²] – [(x – 5)² + (y – 0)²]
= (x² + 10x + 25 + y²) – (x² – 10x + 25 + y²) = – 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 + y²] = (16 + 5x)²
⇒ 16 [x² + y² + 10x + 25] = (16 + 5x)²
⇒ 16x² + 16y² + 160x + 400 = 256 + 160x + 25²
⇒ 9x² + 16y² + 144 = 0
⇒ 9x² – 16y² = 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
PA² – PB² = [(x – 4)² + y²] – [(x + 4)² + y²]
= (x² – 8x + 16 + y²) – (x² + y² + 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
⇒ PA² = (2 – 2x)²
⇒ (x – 4)² + y² = 4 – 8x + 4x²
⇒ x² + y² – 8x + 16 = 4x² – 8x + 4
⇒ 3x² – y² = 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
PA² – PB² = [(x – 0)² + (y – 2)²] – [(x – 0)² + (y + 2)²]
= x² + y² – 4y + 4 – x² – y² – 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)

⇒ 9x² + 9y² + 36 = 81 + 4y²
⇒ 9x² + 5y² = 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)
PA² – PB² = [(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
∴ PA² – PB² = – 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
⇒ 16PA² = (16 – 3y)²
⇒ 16 [(x – 2)² + (y – 3)²] = (16 – 3y)²
⇒ 16 [x² + y² – 4x – 6y + 13] = 256 – 96y + 9y²
⇒ 16x² + 7y² – 64x – 48 = 0
∴ Equation to the locus of P is
16x² + 7y² – 64x – 48 = 0
(or) 16(x² – 4x) + 7y² = 48
⇒ 16 (x² – 4x + 4) + 7y² = 112
⇒ 16 (x – 2)² + 7y² = 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
⇒ \(\frac{1}{2}\) |[x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)[|
= \(\frac{1}{2}\) |[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 ⇒ 9PA² = 4PB²
⇒ 9 [(x – 2)² + (y – 3)²] = 4 [(x – 2)² + (y + 3)²]
⇒ 9 [(x² – 4x + 4 + y² – 6y + 9)] = 4 [x² – 4x + 4 + y² + 6y + 9]
⇒ 5x² + 5y² – 20x – 78y + 65 = 0
∴ Equation to the locus of P is
5x² + 5y² – 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 PA² + PB² = 2PC². 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 PA² + PB² = 2PC²
⇒[(x – 1)² + (y – 2)²] + [(x – 2)² + (y + 3)²]
= 2 [(x + 2)² + (y – 3)²]
⇒ 2x² + 2y² – 6x + 2y + 18
= 2x² + 2y² + 8x – 12y + 26
⇒ 14x – 14y + 8 = 0
⇒ 7x – 7y + 4 = 0
∴ Equation to the locus of P is
7x – 7y + 4 = 0