TS Inter First Year Maths 1B Locus Important Questions

Students must practice these Maths 1B Important Questions TS Inter First Year Maths 1B Locus Important Questions to help strengthen their preparations for exams.

TS Inter First Year Maths 1B Locus Important Questions

Question 1.
Find the equation of the locus of a point that is at a distance 5 from (-2, 3), in the XOY plane. [Mar. ’02; May ’95]
Solution:
Let A (-2, 3) be the given point.
Let P(x, y) be any point on the locus.
The given geometric condition is PA = 5

Squaring on both sides
⇒ 13 + x² + y² + 4x – 6y = 25
⇒ x² + y² + 4x – 6y = 25 – 13
⇒ x² + y² + 4x – 6y = 12
∴ The equation of locus of P(x, y) is x² + y² + 4x – 6y = 12.

Question 2.
Find the equation of the locus of a point P such that the distance of P from the origin is twice the distance of P from A(1, 2). [Mar. ’12; May ’05, ’97]
Solution:
O(0, 0), A(1, 2) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is OP = 2PA

⇒ x² + y² = 4[(x – 1)² + (y – 2)²]
⇒ x² + y² = 4(x² – 2x + 1 + y² – 4y + 4)
⇒ x² + y² = 4x² – 8x + 4 + 4y² – 16y + 16
⇒ 3x² + 3y² – 8x – 16y + 20 = 0
∴ The equation of the locus of P(x, y) is 3x² + 3y² – 8x – 16y + 20 = 0.

Question 3.
Find the equation of the locus of a point equidistant from A(2, 0) and the Y-axis. (May ’03)
Solution:
A(2, 0) is the given point.
Let P(x, y) be any point on the locus.

The distance of P from Y-axis = PN = |x|
The given geometric condition is PA = |PN|
\(\sqrt{(x-2)^2+(y-0)^2}=|x|\)
Squaring on both sides
⇒ \(\left(\sqrt{(x-2)^2+(y-0)^2}\right)^2=(|x|)^2\)
⇒ (x – 2)² + (y)² = x²
⇒ x² – 4x + 4 + y² = x²
⇒ y² – 4x + 4 = 0
∴ The equation of the locus of P(x, y) is y² – 4x + 4 = 0.

Question 4.
Find the equation of the locus of a point P, the square whose distance from the origin is 4 times its y-coordinate. [Mar. ’00]
Solution:
O(0, 0) is the origin.
Let P(x, y) be any point on the locus.
The distance of P from the X-axis = y.

The given geometric condition is OP² = 4 PM
⇒ \(\left(\sqrt{(x-0)^2+(y-0)^2}\right)^2=4 y\)
⇒ x² + y² = 4y
⇒ x² + y² – 4y = 0
∴ The equation of the locus of P(x, y) is x² + y² – 4y = 0.

Question 5.
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|. [Mar. ’00]
Solution:
A = (a, 0), B = (-a, 0) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA² + PB² = 2c²
⇒ \(\left[\sqrt{(x-a)^2+(y-0)^2}\right]^2\) + \(\left[\sqrt{(x+a)^2+(y-0)^2}\right]^2\) = 2c²
⇒ (x – a)² + y² + (x + a)² + y² = 2c²
⇒ x² – 2ax + a² + y² + x² + 2ax + a² + y² = 2c²
⇒ 2x² + 2y² + 2a² = 2c²
⇒ x² + y² + a² = c²
⇒ x² + y² = c² – a²
∴ The equation of the locus of P(x, y) is x² + y² = c² – a²

Question 6.
Find the equation of locus of P, if the line segment joining (2, 3) and (-1, 5) subtends a right angle at P. [Mar. ’13, ’05, ’03; May ’12, ’04, ’03, ’02]
Solution:
Let A(2, 3), B(-1, 5) are the given points.
Let P(x, y) be any point on the locus
The given geometric condition is ∠APB = 90°

Then, PA² + PB² = AB²
⇒ \(\left[\sqrt{(x-2)^2+(y-3)^2}\right]^2+\left[\sqrt{(x+1)^2+(y-5)^2}\right]^2\) = \(\left[\sqrt{(2+1)^2+(3-5)^2}\right]^2\)
⇒ (x – 2)² + (y – 3)² + (x + 1)² + (y – 5)² = (2 + 1)² + (3 – 5)²
⇒ x² + 4 – 4x + y² + 9 – 6y + x² + 1 + 2x + y² + 25 – 10y = 9 + 4
⇒ 2x² + 2y² – 2x – 16y + 26 = 0
⇒ x² + y² – x – 8y + 13 = 0
∴ The equation of the locus of P(x, y) is x² + y² – x – 8y + 13 = 0.

Question 7.
Find the equation of the locus of P, if A = (4, 0), B = (-4, 0) and |PA – PB| = 4. [May ’13]
Solution:
A = (4, 0), B = (-4, 0) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is |PA – PB| = 4
⇒ PA – PB = ±4
⇒ PA = ±4 + PB
Squaring on both sides
⇒ (PA)² = (±4 + PB)²
⇒ PA² = 16 + PB² ± 8PB
⇒ (x – 4)² + (y – 0)² = 16 + (x + 4)² + (y – 0)² ± 8PB
⇒ x² – 8x + 16 + y² = 16 + x² + 8x + 16y² ± 8PB
⇒ -8x = 8x + 16 ± 8PB
⇒ -8x – 8x – 16 = ±8PB
⇒ -16 – 16x = ±8PB
⇒ -2 – 2x = ±PB
Squaring on both sides
⇒ (-2 – 2x)² = (±PB)²
⇒ 4 + 8x + 4x² = PB²
⇒ 4 + 4x² + 8x = (x + 4)² + (y – 0)²
⇒ 4 + 4x² + 8x = x² + 8x + 16 + y²
⇒ 3x² – y² – 12 = 0
∴ The equation of the locus of P(x, y) is 3x² – y² – 12 = 0.

Question 8.
Find the equation of the locus of a point, the sum of whose distances from (0, 2) and (0, -2) is 6. [Mar. ’16 (TS)]
Solution:
Let A = (0, 2), B = (0, -2) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA + PB = 6
⇒ PA = 6 – PB
Squaring on both sides
⇒ PA² = (6 – PB)²
⇒ PA² = 36 – 12PB + PB²
⇒ (x – 0)² + (y – 2)² = 36 + (x – 0)² + (y + 2)² – 12PB
⇒ x² + y² + 4 – 4y = 36 + x² + y² + 4 + 4y – 12PB
⇒ -4y = 36 + 4y – 12PB
⇒ 12PB = 36 + 4y + 4y
⇒ 12PB = 36 + 8y
⇒ 3PB = 9 + 2y
Squaring on both sides
⇒ (3PB)² = (9 + 2y)²
⇒ 9PB² = (9 + 2y)²
⇒ 9[(x – 0)² + (y + 2)²] = 81 + 36y + 4y²
⇒ 9[x² + y² + 4y + 4] = 81 + 36y + 4y²
⇒ 9x² + 9y² + 36y + 36 = 81 + 36y + 4y²
⇒ 9x² + 9y² + 36 – 81 – 4y² = 0
⇒ 9x² + 5y² – 45 = 0
∴ The equation of the locus of P(x, y) is 9x² + 5y² – 45 = 0.

Question 9.
A(2, 3) and B(-3, 4) are two given points. Find the equation of the locus of P, so that the area of the triangle PAB is 8.5. [Mar. ’11]
Solution:
A(2, 3), B(-3, 4) are the points given.
Let P(x, y) be any point on the locus.
The given geometric condition is an area of the triangle, PAB = 8.5

∴ The equation of the locus of P(x, y) is (x + 5y) (x + 5y – 34) = 0
⇒ x² + 5xy – 34x + 5xy + 25y² – 170y = 0
⇒ x² + 10xy + 25y² – 34x – 170y = 0

Question 10.
Find the equation of the locus of p, if the ratio of the distances from p to A(5, -4) and B(7, 6) is 2 : 3. [Mar. ’14, ’98; May ’08, ’01]
Solution:
Let A(5, -4), B(7, 6) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA : PB = 2 : 3
⇒ \(\frac{\mathrm{PA}}{\mathrm{PB}}=\frac{2}{3}\)
⇒ 3PA = 2PB

⇒ 9[x² – 10x + 25 + y² + 16 + 8y] = 4[x² – 14x + 49 + y² – 12y + 36]
⇒ 9x² – 90x + 225 + 9y² + 144 + 72y = 4x² – 56x + 196 – 4y² – 48y + 144
⇒ 9x² – 90x + 225 + 9y² + 144 + 72y – 4x² + 56x – 196 – 4y² + 48y – 144 = 0
⇒ 5x² + 5y² – 34x + 120y + 29 = 0
∴ The equation of the locus of P(x, y) is 5x² + 5y² – 34x + 120y + 29 = 0

Question 11.
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. [Mar. ’19 (T.S); Mar. ’17 (AP); May ’15 (A.P); ’07]
Solution:
A(1, 2), B(2, -3), C(-2, 3) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA² + PB² = 2PC²
⇒ [(x – 1)² + (y – 2)²] + [(x – 2)² + (y + 3)²] = 2[(x + 2)² + (y – 3)²]
⇒ x² – 2x + 1 + y² + 4 – 4y + x² + 4 – 4x + y² + 9 + 6y = 2[x² + 4 + 4x + y² + 9 – 6y]
⇒ 2x² + 2y² – 6x + 2y + 18 = 2x² + 8 + 8x + 2y² + 18 – 12y
⇒ -6x + 2y = 8 + 8x – 12y
⇒ 8 + 8x – 12y + 6x – 2y = 0
⇒ 14x – 14y + 8 = 0
⇒ 7x – 7y + 4 = 0
∴ The equation of the locus of P(x, y) is 7x – 7y + 4 = 0.

Some More Maths 1B Locus Important Questions

Question 12.
Find the equation of the locus of a point that is at a distance 5 from A(4, -3).
Solution:
A(4, -3) is the given point.
Let P(x, y) be any point on the locus.

The given geometric condition is PA = 5
⇒ \(\sqrt{(x-4)^2+(y+3)^2}\) = 5
Squaring on both sides
⇒ \(\left(\sqrt{(x-4)^2+(y+3)^2}\right)^2\) = (5)²
⇒ (x – 4)² + (y + 3)² = 25
⇒ x² – 8x + 16 + y2 + 9 + 6y = 25
⇒ x² + y2 – 8x + 6y = 0
∴ The equation of locus of P(x, y) is x² + y² – 8x + 6y = 0

Question 13.
Find the equation of the locus of a point P, if the distance of P from A(3, 0) is twice the distance of P from B(-3, 0).
Solution:
A(3, 0), B(-3, 0) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA = 2PB

⇒ (x – 3)² + (y – 0)² = 4[(x + 3)² + (y – 0)²]
⇒ x² – 6x + 9 + y² = 4[x² + 6x + 9 + y²]
⇒ x² – 6x + 9 + y² = 4×2 + 24x + 36 + 4y²
⇒ 4x² + 24x + 36 + 4y2 – x² + 6x – 9 – y² = 0
⇒ 3x² + 3y² + 30x + 27 = 0
⇒ x² + y² + 10x + 9 = 0
∴ The equation of the locus of P(x, y) is x² + y² + 10x + 9 = 0.

Question 14.
The ends of the hypotenuse of a right-angled triangle are (0, 6) and (6, 0). Find the equation of the locus of its third vertex. [June ’10; Mar. ’08; Sept. ’00]
Solution:
Let A(0, 6), B(6, 0) are the given points.
Let P(x, y) be the third vertex

The given geometric condition is ∠APB = 90°
⇒ PA² + PB² = AB²
⇒ \(\left[\sqrt{(x-0)^2+(y-6)^2}\right]^2+\left[\sqrt{(x-6)^2+(y-0)^2}\right]^2\) = \(\left[\sqrt{(0-6)^2+(6-0)^2}\right]^2\)
⇒ x² + (y – 6)² + (x – 6)² + y² = (-6)² + (6)²
⇒ x² + y² – 12y + 36 + x² – 12x + 36 + y² = 36 + 36
⇒ 2x² + 2y² – 12x – 12y + 72 = 72
⇒ 2x² + 2y² – 12x – 12y = 0
⇒ x² + y² – 6x – 6y = 0
∴ The equation of the locus of P(x, y) is x² + y² – 6x – 6y = 0.

Question 15.
Find the locus of the third vertex of a right-angled triangle, the ends of whose hypotenuse are (4, 0) and (0, 4). [Mar. ’13(old); Mar. ’18 (TS)]
Solution:
Let A(4, 0), B(0, 4) are the given points.
Let P(x, y) be the third vertex.
The given geometric condition is ∠APB = 90°
⇒ PA² + PB² = AB²

⇒ (x – 4)² + y² + x² + (y – 4)² = (4)² + (-4)²
⇒ x² – 8x + 16 + y² + x² + y² – 8y + 16 = 16 + 16
⇒ 2x² + 2y² – 8x – 8y = 0
⇒ x² + y² – 4y – 4x = 0
⇒ x² + y² – 4x – 4y = 0
∴ The equation of the locus of P(x, y) is x² + y² – 4x – 4y = 0.

Question 16.
Find the equation of the locus of a point, the difference of whose distances from (-5, 0) and (5, 0) is 8.
Solution:
Let A(-5, 0), B(5, 0) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is |PA – PB| = 8
⇒ PA – PB = ±8
⇒ PA = ±8 + PB
Squaring on both sides
⇒ PA² = (±8 + PB)²
⇒ PA² = (±8)² + PB² + 2(±8) PB = 64 + PB² ± 16PB
⇒ (x + 5)² + (y – 0)² = 64 + (x – 5)² + (y – 0)² ± 16PB
⇒ x² + 25 + 10x + y² = 64 + x² – 10x + 25 + y² ± 16PB
⇒ 10x = 64 – 10x ± 16PB
⇒ 10x – 64 + 10x = ±16PB
⇒ 20x – 64 = ±16PB
⇒ 5x – 16 = ±4PB
Squaring on both sides
⇒ (5x – 16)² = (±4PB)²
⇒ 25x² + 256 – 160x = 16PB²
⇒ 25x² + 256 – 160x = 16[(x – 5)² + (y – 0)²]
⇒ 25x² + 256 – 160x = 16[x² + 25 – 10x + y²]
⇒ 25x² + 256 – 160x = 16x² + 400 – 160x + 16y²
⇒ 25x² + 256 – 16x² – 400 – 16y² = 0
⇒ 9x² – 16y² – 144 = 0
⇒ 9x² – 16y² = 144
∴ The equation of the locus of P(x, y) is 9×2 – 16y2 = 114.

Question 17.
Find the equation of the locus of P, if A = (2, 3), B = (2, -3), and PA + PB = 8. [Mar. ’08, ’03]
Solution:
Let A = (2, 3) and B = (2, -3) are the given points..
Let P(x, y) be any point on the locus.
The given geometric condition is PA + PB = 8
⇒ PA = 8 – PB
Squaring on both sides
⇒ PA² = (8 – PB)²
⇒ PA² = 64 – 16PB + PB²
⇒ (x – 2)² + (y – 3)² = (x – 2)² + (y + 3)² + 64 – 16PB
⇒ x² – 4x + 4 + y² – 6y + 9 = x² – 4x + 4 + y² + 6y + 9 + 64 – 16PB
⇒ -6y = 6y + 64 – 16PB
⇒ 16PB = 6y + 6y + 64
⇒ 16PB = 12y + 64
⇒ 4PB = 3y + 16
Squaring on both sides
⇒ 16PB² = (3y + 16)²
⇒ 16PB² = 9y² + 256 + 96y
⇒ 16[(x – 2)² + (y + 3)²] = 9y² + 256 + 96y
⇒ 16[x² – 4x + 4 + y² + 9 + 6y] = 9y² + 256 + 96y
⇒ 16x² – 64x + 64 + 16y² + 144 + 96y – 9y² – 256 – 96y = 0
⇒ 16x² + 7y² – 64x – 48 = 0
∴ The equation of the locus of point P(x, y) is 16x² + 7y² – 64x – 48 = 0.

Question 18
A(5, 3) and B(3, -2) are two fixed points. Find the equation of the locus of P, so that the area of triangle PAB is 9. [Mar. ’09, ’06]
Solution:
Let A(5, 3), B(3, -2) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is Area of the triangle PAB = 9
⇒ \(\frac{1}{2}\left|\begin{array}{ll}
x_1-x_2 & y_1-y_2 \\
x_1-x_3 & y_1-y_3
\end{array}\right|\) = 9
⇒ \(\frac{1}{2}\left|\begin{array}{ll}
x-5 & y-3 \\
x-3 & y+2
\end{array}\right|\) = 9
⇒ \(\left|\begin{array}{ll}
x-5 & y-3 \\
x-3 & y+2
\end{array}\right|\) = 18
⇒ |(x – 5)(y + 2) – (x – 3) (y – 3)| = 18
⇒ |xy – 5y + 2x – 10 – xy + 3x + 3y – 9| = 18
⇒ |5x – 2y – 19| = 18
⇒ (5x – 2y – 19) = ±18
(-) 5x – 2y – 19 = -18
5x – 2y – 19 + 18 = 0
5x – 2y – 1 = 0
(+) 5x – 2y – 19 = 18
5x – 2y – 19 – 18 = 0
5x – 2y – 37 = 0
∴ The equation of the locus of point P(x, y) is (5x – 2y – 1) (5x – 2y – 37) = 0
⇒ 25x² – 10xy – 185x – 10xy + 4y² + 74y – 5x + 2y + 37 = 0
⇒ 25x² – 20xy + 4y² – 190x + 76y + 37 = 0

Question 19.
Find the equation of the locus of a point, which forms a triangle of area 2 with the points A(1, 1) and B(-2, 3). [May ’13(old)]
Solution:
Let A(1, 1), B(-2, 3) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is Area of the triangle PAB = 2
⇒ \(\frac{1}{2}\left|\begin{array}{ll}
x_1-x_2 & y_1-y_2 \\
x_1-x_3 & y_1-y_3
\end{array}\right|\) = 2
⇒ \(\frac{1}{2}\left|\begin{array}{ll}
x-1 & y-1 \\
x+2 & y-3
\end{array}\right|\) = 2
⇒ |(x – 1)(y – 3) – (y – 1) (x + 2)| = 4
⇒ |xy – 3x – y + 3 – xy – 2y + x + 2| = 4
⇒ |-2x – 3y + 5 | = 4
⇒ -2x – 3y + 5 = ±4
(+) -2x – 3y + 5 = 4
2x + 3y – 5 + 4 = 0
2x + 3y – 1 = 0
(-) -2x – 3y + 5 = -4
2x + 3y – 5 – 4 = 0
2x + 3y – 9 = 0
∴ The equation of the locus of P(x, y) is (2x + 3y – 1) (2x + 3y – 9) = 0.
⇒ 4x² + 6xy – 18x + 6xy + 9y² – 27y – 2x – 3y + 9 = 0
⇒ 4x² + 12xy + 9y² – 20x – 30y + 9 = 0

Question 20.
If the distance from p to the points (2, 3) and (2, -3) are in the ratio 2 : 3, then find the equation of the locus of P. [May ’15 (TS)]
Solution:
Let A(2, 3), B(2, -3) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA : PB = 2 : 3

⇒ 9[x² + 4 – 4x + y² + 9 – 6y] = 4[x² + 4 – 4x + y² + 9 + 6y]
⇒ 9x² + 36 – 36x + 9y² + 81 – 54y – 4x² + 16 – 16x + 4y² + 36 + 24y
⇒ 9x² + 36 – 36x + 9y² + 81 – 54y – 4x² – 16 + 16x – 4y² – 36 – 24y = 0
⇒ 5x² + 5y² – 20x – 78y + 65 = 0
∴ The equation of the locus of P(x, y) is 5x² + 5y² – 20x – 78y + 65 = 0.

Question 21.
Find the equation of the locus of a point that is equidistant from the coordinate axes.
Solution:
P(x, y) is any point on the locus.
Let M, and N are the projections drawn from P to the X-axis and Y-axis respectively.
The distance of P from the X-axis is |PM| = |y|
The distance of P from the Y-axis is |PN| = |x|
The given geometric condition is |PM| = |PN|
|y| = |x|

Squaring on both sides |y| = |x|
⇒ y² = x²
⇒ x² – y² = 0
∴ The equation of the locus of P(x, y) is x2 – y2 = 0.

Question 22.
Find the equation of the locus of a point that is equidistant from points A(-3, 2) and B(0, 4).
Solution:
A(-3, 2) and B(0, 4) are the given points.
Let P(x, y) be any point on the locus.
The given geometric condition is PA = PB

⇒ (x + 3)² + (y – 2)² = (x – 0)² + (y – 4)²
⇒ x² + 6x + 9 + y² – 4y + 4 = x² + y² + 16 – 8y
⇒ 13 + 6x – 4y = 16 – 8y
⇒ 13 + 6x – 4y – 16 + 8y = 0
⇒ 6x + 4y – 3 = 0
∴ The equation of the locus of P(x, y) is 6x + 4y – 3 = 0.