Question

Let $S$ be the focus of the parabola $y^2=8x$ and let $PQ$ be the common chord of the circle $x^2+y^2-2x-4y=0$ and the given parabola. The area of the $△PQS$ is

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is A $4$

Given Parabola:

$\Rightarrow$$y^2=8x$

Given Circle: $(x-1{)}^2+(y-2{)}^2=5$

$\Rightarrow$Take a point on the parabola as $(2t^2,4t)\equiv (x,y)$

Solve equations simultaneously

$\Rightarrow$$4t^4+16t^2-4t^2-16t=0$

$\Rightarrow t^4+3t^2-4t=0\Rightarrow t=0,1$

$\Rightarrow$We get the points as $P(0,0)$ and $Q(2,4)$.

$\Rightarrow$The distance between $(0,0)\equiv (x_1,y_1)$ and $(2,4)\equiv (x_2,y_2)$ is given by,

$\Rightarrow$Distance Formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$\therefore$ Distance$=2\sqrt{5}$

$\Rightarrow$Focus of parabola $y^2=8x$ has coordinates $(2,0)$.

$\Rightarrow$$PQS$ will form a right angled triangle with area $\frac{1}{2}\times 2\times 4=4$ square units.