$y=\sin x \Rightarrow \frac{d y}{d x}=\cos x$
Slope of normal $\Rightarrow m=-\frac{1}{\frac{d y}{d x}}=-\frac{1}{\cos x}$
Let the co-ordinate of point $\mathrm{P}$ is $(h, k)$ Then, $m=-\frac{1}{\cos h}$ Equation of normal, which passes through $(0,0)$ is
$(y-0)=m(x-0) \Rightarrow y=-\frac{1}{\cos h} x$
$\because$ Above equation passes through $(h, k)$
$\therefore k=-\frac{h}{\cos h} \Rightarrow \cos h=-\frac{h}{k}$ ...(i)
Also, $(h, k)$ lies on $y=\sin x$
$\therefore k=\sin h \Rightarrow \sin h=k$ ..(ii)
Equation (i) ${ }^2+(\text { ii })^2$ :
$\begin{aligned} & \Rightarrow \cos ^2 h+\sin ^2 h=\frac{h^2}{k^2}+k^2 \\ & \Rightarrow 1=\frac{h^2+k^4}{k^2} \Rightarrow h^2+k^4=k^2\end{aligned}$
Taking locus of point $(h, k)$, we get
$\begin{aligned} & x^2+y^4=y^2 \\ & \Rightarrow x^2=y^2-y^4\end{aligned}$