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Question

Mathematics Question on Parabola

Let SS be the focus of the parabola y2=8xy ^{2}=8 x and let PQPQ be the common chord of the circle x2+y22x4y=0x^{2}+y^{2}-2 x-4 y=0 and the given parabola. The area of the ΔPQS\Delta PQS is

A

4 sq units

B

3 sq units

C

2 sq units

D

8 sq units

Answer

4 sq units

Explanation

Solution

Given Parabola:
y2=8x\Rightarrow y ^{2}=8 x
Given Circle: (x1)2+(y2)2=5( x -1)^{2}+( y -2)^{2}=5
\Rightarrow Take a point on the parabola as (2t2,4t)(x,y)\left(2 t^{2}, 4 t\right) \equiv(x, y)
Solve equations simultaneously
4t4+16t24t216t=0\Rightarrow 4 t ^{4}+16 t ^{2}-4 t ^{2}-16 t =0
t4+3t24t=0\Rightarrow t ^{4}+3 t ^{2}-4 t =0
t=0,1\Rightarrow t =0,1
\Rightarrow We get the points as P(0,0)P(0,0) and Q(2,4)Q(2,4).
\Rightarrow The distance between (0,0)(x1,y1)(0,0) \equiv\left( x _{1}, y _{1}\right) and (2,4)(x2,y2)(2,4) \equiv\left( x _{2}, y _{2}\right) is given by,
\Rightarrow Distance Formula =(x2x1)2+(y2y1)2=\sqrt{\left( x _{2}- x _{1}\right)^{2}+\left( y _{2}- y _{1}\right)^{2}}
\therefore Distance =25=2 \sqrt{5}
\Rightarrow Focus of parabola y2=8xy^{2}=8 x has coordinates (2,0)(2,0).
PQS\Rightarrow P QS will form a right angled triangle with area 12×2×4=4\frac{1}{2} \times 2 \times 4=4 square units.