$\text{ Let }\mathrm{RS}\text{ be the diameter of the circle }{x}^2+y^2=1,\text{ where }\mathrm{S}\text{ is the point }(1,0)\text{ . Let }\mathrm{P}\text{ be a }$variable point (other than R and S) on the circle and tangents to the circle at S and P meet at the point Q. The normal to the circle at P intersects a line drawn through Q parallel to RS at point E. Then the locus of E passes through the point (s)

$\text{ Let }P(\cos \alpha ,\sin \alpha );s(1,0) and \text{ RS is }X\text{ -axis and Tangent at P is }x\cos \alpha +y\sin \alpha =1$$\to \left(1\right)$

$\tan gent at S\left(1,0\right) is x=1 \to \left(2\right)$

$solve \left(1\right),\left(2\right)$

$\therefore Q\left(1,\frac{1-\cos \alpha }{\sin \alpha }\right) , now \text{ Normal at }\mathrm{P}\text{ is }x\sin \alpha -y\cos \alpha =0 \to \left(3\right)$

$\text{ Line through Q parallel}\text{ to }\mathrm{x}\text{ -axis is }y=\frac{1-\cos \alpha }{\sin \alpha }$  , $solve with \left(3\right)$,

$\therefore E\left(\frac{\cos \alpha }{1+\cos \alpha },\frac{1-\cos \alpha }{\sin \alpha }\right)=(h,k)$

$$\begin{gathered} \text{ we have to find Locus of E,} \\ now k=\frac{1-\cos \alpha }{\sin \alpha }=\tan \frac{\alpha }{2} \end{gathered}$$

$\begin{array}{l}h=1-\frac{1}{1+\cos \alpha }\\ =1-\frac{1}{2}{\sec }^2\frac{\alpha }{2}\\ =1-\frac{1}{2}\left(1+k^2\right)\\ \therefore locusofE\text{\hspace{0.17em} is }x=\frac{1}{2}-\frac{1}{2}{y}^2\\ \\ \therefore y^2=1-2x\end{array}$