At any point on the curve $\frac{\mathrm{a}}{{\mathrm{x}}^2}+\frac{\mathrm{b}}{{\mathrm{y}}^2}=1,$ the $\mathrm{y}=$intercept made by the tangent is proportional to

Let $\mathrm{P}\left({\mathrm{x}}_1 {\mathrm{y}}_1\right)$ lie on the curve 

$\begin{array}{rcl}\frac{\mathrm{a}}{{{\mathrm{x}}_1}^2}+\frac{\mathrm{b}}{{{\mathrm{y}}_1}^2}& =& 1 -\left(1\right)\\ \frac{\mathrm{dy}}{\mathrm{dx}}& =& \frac{{\displaystyle \frac{-2\mathrm{a}}{{\mathrm{x}}^3}}}{{\displaystyle \frac{2\mathrm{b}}{{\mathrm{y}}^3}}}=-\frac{\mathrm{a}}{\mathrm{b}} \frac{{{\mathrm{y}}_1}^3}{{{\mathrm{x}}_1}^3}\end{array}$

eq of Tangent

$\left(\mathrm{a} {{\mathrm{y}}_1}^3\right)\mathrm{x}+\left(\mathrm{b} {{\mathrm{x}}_1}^3\right)\mathrm{y}={{\mathrm{x}}_1}^3 {{\mathrm{y}}_1}^3 \left(\frac{\mathrm{a}}{{{\mathrm{x}}_1}^2}+\frac{\mathrm{b}}{{{\mathrm{y}}_1}^2}\right)$

meet y-axis $\Rightarrow \mathrm{x}=0$

         $\mathrm{y}=\frac{{{\mathrm{y}}_1}^3}{\mathrm{b}}$