Question

Let $PQ$ be the focal chord of the parabola $y^{2}=4x.$ If the center of the circle having $PQ$ as its diameter lies on the line $y=\frac{4}{\sqrt{5}},$ then the radius (in units) is equal to

Answer 1

Let, $P$ is $\left(t^{2} , 2 t\right)$ then,
$Q$ is $\left(\frac{1}{t^{2}} , \frac{- 2}{t}\right)$
Midpoint of $PQ$ is $\left(\frac{t^{2} + \frac{1}{t^{2}}}{2} , \frac{2 t - \frac{2}{t}}{2}\right)$
$\Rightarrow t-\frac{1}{t}=\frac{4}{\sqrt{5}}\Rightarrow t^{2}+\frac{1}{t^{2}}-2=\frac{16}{5}$
$\Rightarrow t^{2}+\frac{1 }{t^{2}}+2=\frac{36}{5}$
$\Rightarrow \left(t^{2} + 1\right)+\left(\frac{1}{t^{2}} + 1\right)=7.2$
If $S$ is the focus, then
$PS+SQ=7.2\Rightarrow PQ=7.2$
Hence, radius $=3.6$ units