Mathematics - Parabola Question with Solution | TestHub

$\mathrm{y}=\mathrm{x}+1$ will be parallel to tangent at

$\mathrm{P}\left({\mathrm{y}}_0^2,{\mathrm{y}}_0\right)$ on $\mathrm{x}={\mathrm{y}}^2$ $\Rightarrow \frac{1}{2{\mathrm{y}}_0}=1\Rightarrow {\mathrm{y}}_0=\frac{1}{2}$

$\Rightarrow \mathrm{p}\left(\frac{1}{4},\frac{1}{2}\right)$ $\Rightarrow {\perp }^{\mathrm{r}}$ distance $=\left|\frac{\frac{1}{4}-\frac{1}{2}+1}{\sqrt{2}}\right|$ $=\frac{\frac{3}{4}}{\sqrt{2}}=\frac{3\sqrt{2}}{8}$