The sum of the squares of the perpendicular drawn from the points (0,…

The equation of any tangent to a curve $y=f(x)$ is

$Y-y=\frac{d y}{d x}(X-x)$ or, $X \frac{d y}{d x}-Y+\left(y-x \frac{d y}{d x}\right)=0$     ....(i)

It is given that

$\left|\frac{-1+y-x \frac{d y}{d x}}{\sqrt{\left(\frac{d y}{d x}\right)^2+1}}\right|^2+\left|\frac{1+y-x \frac{d y}{d x}}{\sqrt{\left(\frac{d y}{d x}\right)^2+1}}\right|^2=2$

$\Rightarrow 2\left\{\left(y-x \frac{d y}{d x}\right)^2+1\right\}=2\left\{1+\left(\frac{d y}{d x}\right)^2\right\}$

$\Rightarrow \left(y-x \frac{d y}{d x}\right)^2=\left(\frac{d y}{d x}\right)^2$

$\Rightarrow y-x \frac{d y}{d x}= \pm \frac{d y}{d x}$

$\Rightarrow \frac{1}{x \pm 1} d x=\frac{1}{y} d y$

$\Rightarrow \log (x \pm 1)=\log y+\log C$

$\Rightarrow \log (x \pm 1)=\log y+\log C$

$\Rightarrow C y=x \pm 1$ or, $y=k(x \pm 1)$