Question

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

• A. $6$ sq units
• B. $16$ sq units
• C. $4$ sq units
• D. $64$ sq units

Answer 1

The correct option is C

$4$ sq units

Explanation of correct answer :

Finding are of $△PQS$.

Equation of parabola, $y^2=8x$

As we know, parametric points for parabola $y^2=4ax$ is $\left(a{t}^2,2at\right)\dots \left(1\right)(\text{where}t=\text{parameter})$

Thus, parametric points for parabola $y^2=8x$ is $\left(2t^2,4t\right)$

Equation of circle, $x^2+y^2-2x-4y=0$

Substitute the value of $x=2t^2$ and $y=4t$ in the equation of circle.

$$\begin{gathered} \Rightarrow {\left(2t^2\right)}^2+{\left(4t\right)}^2-2\left(2t^2\right)-4\left(4t\right)=0 \\ \Rightarrow 4t^4+16t^2-4t^2-16t=0 \\ \Rightarrow 4t^4+12t^2-16t=0 \\ \Rightarrow 4t(t^3+3t-4)=0 \\ \Rightarrow 4t(t-1)(t^2+t+4)=0 \\ t=0\text{and}t=1\text{and}{t}^2+t+4=0 \end{gathered}$$

Now,

$t^2+t+4=0$ has imaginary roots, thus the values of t are $0$ and $1$.

Also, we have been given,

$y^2=4ax$ and $y^2=8x$

So from these, we get,

$$\begin{gathered} 4ax=8x \\ \Rightarrow a=\frac{8x}{4x} \\ \Rightarrow a=2 \end{gathered}$$

Since $a=2$ ,the focus of parabola, $S=\left(2,0\right)$

Putting the values of $t$ in equation $\left(1\right)$, we get points $P\text{and}Q$.

$P=(0,0),Q=(2,4)$

Using points $P,Q,S$ a triangle is formed.

Thus, Area of $△PQS$ is

= $$\begin{gathered} \frac{1}{2}\left|4\times 2-0\right| \\ =\frac{1}{2}\left|8\right| \\ =4 \end{gathered}$$

Hence, the correct answer is option (C).