12 Most SAQ’s of Locus Chapter in Inter 1st Year Maths-1B (TS/AP)
4 Marks
SAQ-1 : Find the equation of locus of P, if the line segment joining (4,0) and (0,4) subtends a right angle at P
Let the locus point P = (x,y)
Given points $$A = (4,0) B = (0,4)$$
Given condition: $$\angle APB = 90^\circ$$
$$\Rightarrow PA^2 + PB^2 = AB^2$$
$$\Rightarrow [(x-4)^2 + (y-0)^2] + [(x-0)^2 + (y-4)^2] = (4-0)^2 + (0-4)^2$$
$$\Rightarrow (x^2 – 8x + 16) + y^2 + x^2 + (y^2 – 8y + 16) = 16 + 16$$
$$\Rightarrow 2x^2 + 2y^2 – 8x – 8y + 32 = 32$$
$$\Rightarrow 2x^2 + 2y^2 – 8x – 8y = 0$$
$$\Rightarrow x^2 + y^2 – 4x – 4y = 0$$
Hence, the locus of P is $$x^2 + y^2 – 4x – 4y = 0$$
SAQ-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
We take A = (0,6) B = (6,0) and P = (x,y) is a point on the locus,
Given condition: $$\angle APB = 90^\circ$$
$$\Rightarrow PA^2 + PB^2 = AB^2$$
$$\Rightarrow [(x-0)^2 + (y-6)^2] + [(x-6)^2 + (y-0)^2] = (0-6)^2 + (6-0)^2$$
$$\Rightarrow x^2 + (y^2 – 12y + 36) + (x^2 – 12x + 36) + y^2 = 36 + 36$$
$$\Rightarrow 2x^2 + 2y^2 – 12x – 12y + 72 = 72$$
$$\Rightarrow 2x^2 + 2y^2 – 12x – 12y = 0$$
$$\Rightarrow x^2 + y^2 – 6x – 6y = 0$$
Hence, the locus of P is $$x^2 + y^2 – 6x – 6y = 0$$
SAQ-3 : Find the equation of locus of P, if the line segment joining (2,3) and (-1,5) subtends a right angle at P
Let the locus point P = (x,y)
Given points $$A = (2,3) B = (-1,5)$$
Given condition: $$\angle APB = 90^\circ$$
$$\Rightarrow PA^2 + PB^2 = AB^2$$
$$\Rightarrow [(x-2)^2 + (y-3)^2] + [(x+1)^2 + (y-5)^2] = (2+1)^2 + (3-5)^2$$
$$\Rightarrow [(x^2 – 4x + 4) + (y^2 – 6y + 9)] + [(x^2 + 2x + 1) + (y^2 – 10y + 25)] = 9 + 4$$
$$\Rightarrow 2x^2 + 2y^2 – 2x – 16y + 39 = 13$$
$$\Rightarrow 2x^2 + 2y^2 – 2x – 16y + 26 = 0$$
$$\Rightarrow x^2 + y^2 – x – 8y + 13 = 0$$
Hence, the locus of P is $$x^2 + y^2 – x – 8y + 13 = 0$$
SAQ-4 : A(5,3) and B(3,-2) are two fixed points. Find the equation of locus of P, so that the area of triangle PAB is 9 sq.units
Let the locus point P = (x,y)
Given points $$A = (5,3) B = (3,-2)$$
Given condition:
Area of $$\Delta PAB = 9 sq.units$$
$$\Delta = \frac{1}{2} \left| \begin{array}{cc}
5 – 3 & 5 – x \
3 + 2 & 3 – y \
\end{array} \right| = 9$$
$$\Rightarrow \frac{1}{2} \left| \begin{array}{cc}
2 & 5 – x \
5 & 3 – y \
\end{array} \right| = 9$$
$$\Rightarrow \left| 2(3 – y) – 5(5 – x) \right| = 18$$
$$\Rightarrow \left| 6 – 2y – 25 + 5x \right| = 18$$
$$\Rightarrow \left| 5x – 2y – 19 \right| = 18$$
$$\Rightarrow 5x – 2y – 19 = \pm 18$$
$$\Rightarrow 5x – 2y – 19 = 18 \text{ (or) } 5x – 2y – 19 = -18$$
$$\Rightarrow 5x – 2y – 37 = 0 \text{ (or) } 5x – 2y – 1 = 0$$
$$\Rightarrow (5x – 2y – 37)(5x – 2y – 1) = 0$$
Hence, the locus of P is $$(5x – 2y – 37)(5x – 2y – 1) = 0$$
SAQ-5 : A(2,3) and B(-3,4) be two given points. Find the equation of the locus of P so that the area of the triangle PAB is 8.5 sq.units
Let the locus point P = (x,y)
Given points $$A = (2,3) B = (-3,4)$$
Given condition:
Area of $$\Delta PAB = 8.5 sq.units$$
$$\Delta = \frac{1}{2} \left| \begin{array}{cc}
2 + 3 & 2 – x \
3 – 4 & 3 – y \
\end{array} \right| = 8.5$$
$$\Rightarrow \frac{1}{2} \left| \begin{array}{cc}
5 & 2 – x \
-1 & 3 – y \
\end{array} \right| = 8.5$$
$$\Rightarrow \left| 5(3 – y) + 1(2 – x) \right| = 17$$
$$\Rightarrow \left| 15 – 5y + 2 – x \right| = 17$$
$$\Rightarrow \left| 17 – 5y – x \right| = 17$$
$$\Rightarrow 17 – 5y – x = \pm 17$$
$$\Rightarrow 17 – 5y – x = 17 \text{ (or) } 17 – 5y – x = -17$$
$$\Rightarrow -5y – x = 0 \text{ (or) } -5y – x = -34$$
$$\Rightarrow x + 5y = 0 \text{ (or) } x + 5y – 34 = 0$$
$$\Rightarrow (x + 5y)(x + 5y – 34) = 0$$
Hence, the locus of P is $$(x + 5y)(x + 5y – 34) = 0$$
SAQ-6 : A(1,2), B(2,-3), C(-2,3) are 3 points. A point P moves such that PA2 + PB2 = 2PC2. Show that the equation to the locus of P is 7x – 7y + 4 = 0
Let the locus point P = (x,y)
Given points are $$A = (1,2) B = (2,-3) C = (-2,3)$$
Given condition: $$PA^2 + PB^2 = 2PC^2$$
$$\Rightarrow [(x-1)^2 + (y-2)^2] + [(x-2)^2 + (y+3)^2] = 2[(x+2)^2 + (y-3)^2]$$
$$\Rightarrow (x^2 + 1 – 2x) + (y^2 + 4 – 4y) + (x^2 + 4 – 4x) + (y^2 + 9 + 6y) = 2[(x^2 + 4 + 4x) + (y^2 + 9 – 6y)]$$
$$\Rightarrow 2x^2 + 2y^2 – 6x + 2y + 18 = 2x^2 + 2y^2 + 8x – 12y + 26$$
$$\Rightarrow -6x – 8x + 2y + 12y + 18 – 26 = 0$$
$$\Rightarrow -14x + 14y – 8 = 0$$
$$\Rightarrow -2(7x – 7y + 4) = 0$$
$$\Rightarrow 7x – 7y + 4 = 0$$
Hence, the locus of P is $$7x – 7y + 4 = 0$$
SAQ-7 : Find the equation of locus of a point P, if A = (2,3), B = (2,-3) and PA + PB = 8
Given A = (2,3) B = (2,-3) and P = (x,y) is a point on the locus with the condition that
$$PA + PB = 8$$
$$PA^2 = (8 – PB)^2$$
$$PA^2 = 64 + PB^2 – 16PB$$
$$16PB = 64 + [(x-2)^2 + (y+3)^2] – [(x-2)^2 + (y-3)^2]$$
$$16PB = 64 + (y+3)^2 – (y-3)^2$$
$$16PB = 64 + 4(3)y$$
$$16PB = 4(16 + 3y)$$
$$4PB = 16 + 3y$$
Squaring on both sides:
$$16PB^2 = (16 + 3y)^2$$
$$16[(x – 2)^2 + (y + 3)^2] = 256 + 9y^2 + 96y$$
$$16[(x^2 + 4 – 4x) + (y^2 + 9 + 6y)] = 256 + 9y^2 + 96y$$
$$16x^2 + 64 – 64x + 16y^2 + 144 + 96y – 256 – 9y^2 – 96y = 0$$
$$16x^2 + 7y^2 – 64x – 48 = 0$$
Hence, the locus of P is $$16x^2 + 7y^2 – 64x – 48 = 0$$
SAQ-8 : Find the equation of locus of a point the sum of whose distances from (0,2),(0,-2) is 6 units
Given A = (0,2) B = (0,-2) and P(x,y) is a point on the locus with the condition that
$$PA + PB = 6$$
$$PA = 6 – PB$$
$$PA^2 = (6 – PB)^2$$
$$PA^2 = 36 + PB^2 – 12PB$$
$$12PB = 36 + PB^2 – PA^2$$
$$12PB = 36 + [(x – 0)^2 + (y + 2)^2] – [(x – 0)^2 + (y – 2)^2]$$
$$12PB = 36 + (y + 2)^2 – (y – 2)^2$$
$$12PB = 36 + 4(2)y$$
$$12PB = 4(9 + 2y)$$
$$3PB = 9 + 2y$$
Squaring on both sides:
$$9PB^2 = (9 + 2y)^2$$
$$9[(x – 0)^2 + (y + 2)^2] = 81 + 4y^2 + 36y$$
$$9(x^2 + y^2 + 4y + 4) = 81 + 4y^2 + 36y$$
$$9x^2 + 9y^2 + 36y + 36 = 81 + 4y^2 + 36y$$
$$9x^2 + 5y^2 = 45$$
Hence, the locus of P is $$9x^2 + 5y^2 = 45$$
SAQ-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
Given A = (2,3) B = (2,-3) and P = (x,y) is a point on the locus with the condition
$$\frac{PA}{PB} = \frac{2}{3}$$
$$3PA = 2PB$$
$$9PA^2 = 4PB^2$$
$$9[(x – 2)^2 + (y – 3)^2] = 4[(x – 2)^2 + (y + 3)^2]$$
$$9[(x^2 + 4 – 4x) + (y^2 + 9 – 6y)] = 4[(x^2 + 4 – 4x) + (y^2 + 9 + 6y)]$$
$$9x^2 + 36 – 36x + 9y^2 + 81 – 54y = 4x^2 + 16 – 16x + 4y^2 + 36 + 24y$$
$$9x^2 – 4x^2 + 9y^2 – 4y^2 – 36x + 16x – 54y – 24y + 81 – 16 = 0$$
$$5x^2 + 5y^2 – 20x – 78y + 65 = 0$$
Hence, the locus of P is $$5x^2 + 5y^2 – 20x – 78y + 65 = 0$$
SAQ-10 : Find the equation of locus of a point the difference of whose distances from (-5,0) and (5,0) is 8 units
Given A = (-5,0) B = (5,0) and
P = (x,y) is a point on the locus
Given condition: $$|PA – PB| = 8$$
$$\Rightarrow PA – PB = \pm 8$$
$$\Rightarrow PA = \pm8 + PB$$
$$\Rightarrow PA^2 = (\pm8 + PB)^2$$
$$\Rightarrow PA^2 = 64 + PB^2 \pm 16PB$$
$$\Rightarrow \pm16PB = 64 + PB^2 – PA^2$$
$$\Rightarrow \pm16PB = 64 + [(x – 5)^2 + (y – 0)^2]] – [(x + 5)^2 + (y – 0)^2)]$$
$$\Rightarrow \pm16PB = 64 + (x – 5)^2 – (x + 5)^2$$
$$\Rightarrow \pm16PB = 64 – 4(x)(5)$$
$$\Rightarrow \pm16PB = 64 – 20x$$
$$\Rightarrow \pm16PB = 4(16 – 5x)$$
$$\Rightarrow \pm4PB = 16 – 5x$$
Squaring on both sides
$$\Rightarrow 16PB^2 = (16 – 5x)^2$$
$$\Rightarrow 16[(x – 5)^2 + (y – 0)^2] = 256 + 25x^2 – 160x$$
$$\Rightarrow 16(x^2 + 25 – 10x + y^2) = 256 – 160x + 25x^2$$
$$\Rightarrow 16x^2 + 400 – 160x + 16y^2 = 256 – 160x + 25x^2$$
$$\Rightarrow (25x^2 – 16x^2) – 16y^2 = 400 – 256$$
$$\Rightarrow 9x^2 – 16y^2 = 144$$
Hence locus of P is $$9x^2 – 16y^2 = 144$$
SAQ-11 : Find the equation of locus P id A = (4,0), B = (-4,0) and |PA – PB| = 4
Given A = (4,0) B = (-4,0) and P = (x,y) is a point on the locus
Given condition: $$|PA – PB| = 4$$
$$\Rightarrow PA – PB = \pm4$$
$$\Rightarrow PA = \pm4 + PB$$
$$\Rightarrow PA^2 = (\pm4 + PB)^2$$
$$\Rightarrow PA^2 = 16 + PB^2 \pm 8PB$$
$$\Rightarrow \pm8PB = 16 + PB^2 – PA^2$$
$$\Rightarrow \pm8PB = 16 + PB^2 – PA^2$$
$$\Rightarrow \pm8PB = 16 + [(x + 4)^2 + (y – 0)^2] – [(x – 4)^2 + (y – 0)^2]$$
$$\Rightarrow \pm8PB = 16 + (x + 4)^2 – (x – 4)^2$$
$$\Rightarrow \pm8PB = 16 + 4(x)(4)$$
$$\Rightarrow \pm8PB = 16 + 16x$$
$$\Rightarrow \pm8PB = 16(x + 1)$$
Squaring on both sides
$$\Rightarrow 64[(x + 4)^2 + (y – 0)^2] = 256(x^2 + 2x + 1)$$
$$\Rightarrow (x^2 + 16 + 8x) + y^2 = 4x^2 + 4 + 8x$$
$$\Rightarrow 3x^2 – y^2 – 12 = 0$$
Hence, the locus of P is $$3x^2 – y^2 – 12 = 0$$
SAQ-12 : Find the equation of locus of a point P, if the distance of P from A(3,0) is twice the distance of P from B(-3,0)
Let P(x,y) be a point on the locus
Given condition: $$PA = 2PB$$
$$\Rightarrow PA^2 = 4PB^2$$
$$\Rightarrow (x – 3)^2 + y^2 = 4[(x + 3)^2 + y^2]$$
$$\Rightarrow x^2 – 6x + 9 + y^2 = 4[x^2 + 6x + 9 + y^2]$$
$$\Rightarrow x^2 – 6x + 9 + y^2 = 4x^2 + 24x + 36 + 4y^2$$
$$\Rightarrow 3x^2 + 3y^2 + 30x + 27 = 0$$
$$\Rightarrow 3(x^2 + y^2 + 10x + 9) = 0$$
$$\Rightarrow x^2 + y^2 + 10x + 9 = 0$$
The equation of the locus of P is $$x^2 + y^2 + 10x + 9 = 0$$