Question

Let $N$ be the normal on $y^2=4x$ at $S(1,2)$. A circle is inscribed on $SP$ as diameter, where $P$ is the focus of $y^2=4x$. If the length of the intercept made by the circle on $N$ is $k$, then the value of $k^4$ is

Answer 1

$y^2=4x$
Focus is $P(1,0)$
Equation of the circle with $S(1,2)$ and $P(1,0)$ as extremities of diameter is,
$(x-1)(x-1)+(y-2)(y-0)=0\Rightarrow (x-1{)}^2+(y-1{)}^2=1$
Centre and radius of the circle is, $C(1,1),r=1$

$y^2=4x$
$\Rightarrow 2y{y}^{\prime }=4$
Slope of the normal at $S(1,2)$ is,
$m=-\frac{1}{y^{\prime }}=-\frac{y}{2}=-1$
Hence, equation of the normal is,
$y-2=-1(x-1)\Rightarrow x+y-3=0$

Let the distance of the normal from the centre of the circle be $d$.
$d=\left|\frac{1+1-3}{\sqrt{2}}\right|=\frac{1}{\sqrt{2}}$
Length of the intercept made by the circle on the normal
$=2\sqrt{r^2-d^2}=2\sqrt{1-\frac{1}{2}}=\sqrt{2}\therefore k^4=4$