Question

The common tangent to the circle $x^2+y^2=9$ and the parabola $y^2=8x$ touch the circle at the points $P,Q$ and the parabola at the points $R,S$. Then the area of the quadrilateral $PQRS$ is

Answer 1

We have,

Given that,

Equation of circle is $x^2+y^2=9$ …… (1)

Equation of parabola is $y^2=8x......\left(2\right)$

Let the length of quadrilateral $=x$ and Width $=y$

From equation (1) and (2) to, and we get,

$x^2+y^2=9$

$\Rightarrow x^2+8x-9=0$

Using quadratic formula and we get,

$x=\frac{-8\pm \sqrt{8^2-4\times 1\times (-9)}}{2\times 1}$

$x=\frac{-8\pm \sqrt{64+36}}{2}$

$x=\frac{-8\pm \sqrt{100}}{2}$

$x=\frac{-8\pm 10}{2}$

On taking +ive sign,

$x=\frac{-8+10}{2}$

$x=1$

Now,

Taking –ive sign and we get,

$x=\frac{-8-10}{2}$

$x=-9\left(\text{-ive}\text{not}\text{exists}\right)$

Put the value of x in (2) and we get,

$y^2=8x$

$\Rightarrow y^2=8\times 1$

$\Rightarrow y=2\sqrt{2}$

Then,

Area of quadrilateral $=l\times b$

$=1\times 2\sqrt{2}$

$=2\sqrt{2}sq.unit$

Hence, this is the answer.