Question

The sum of squares of the length of the perpendiculars drawn from the points $(0,1)$ and $(0,-1)$ to any tangent to a curve $y=f\left(x\right)$ is $2$ units. Then the number of roots of $f\left(x\right)=0$ is

Answer 1

Equation of tangent to the curve $y=f\left(x\right)$ at $(x,y)$ is
$Y-y=\frac{dy}{dx}(X-x)$
The length of the perpendicular from $(0,1)$ to the tangent is,
$D_1=\left|\frac{1+x{y}^{\prime }-y}{\sqrt{1+(y^{\prime }{)}^2}}\right|$
where $y^{\prime }=\frac{dy}{dx}$
The length of the perpendicular from $(0,-1)$ to the tangent is,
$D_2=\left|\frac{-1+x{y}^{\prime }-y}{\sqrt{1+(y^{\prime }{)}^2}}\right|$

Given, $D_1^2+D_2^2=2$
$\Rightarrow 2+2x^2{y}^{\prime 2}+2y^2-4xy{y}^{\prime }=2(1+y^{\prime 2})\Rightarrow x^2{y}^{\prime 2}+y^2-2xy{y}^{\prime }=y^{\prime 2}\Rightarrow {\left(\frac{dy}{dx}\right)}^2(x^2-1)-2xy\left(\frac{dy}{dx}\right)+y^2=0\Rightarrow \frac{dy}{dx}=\frac{2xy\pm \sqrt{4y^2}}{2(x^2-1)}=\frac{xy\pm y}{x^2-1}\Rightarrow \frac{dy}{y}=\frac{(x\pm 1)dx}{x^2-1}$

Case $1:$ Taking positive sign
$\frac{dy}{y}=\frac{dx}{x-1}$
$\Rightarrow y=e^{C_1}(x-1)$
Case $2:$ Taking negative sign
$\frac{dy}{y}=\frac{dx}{x+1}$
$\Rightarrow y=e^{C_2}(x+1)$
So, $f\left(x\right)=0$ will have two roots i.e., $x=-1,1$