Mathematics - Application of Derivative Question with Solution | TestHub

Slope of $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $P\left(x_0,y_0\right)$ is $-\frac{b^2{x}_0}{a^2{y}_0}$; slope of $\frac{x^2}{{\alpha }^2}-\frac{y^2}{{\beta }^2}=1$ at $P\left(x_0,y_0\right)$ is $\frac{{\beta }^2{x}_0}{{\alpha }^2{y}_0}$ $\because m_1{m}_2=-1\Rightarrow b^2{\beta }^2{x}_0^2=a^2{\alpha }^2{y}_0^2\dots (1)$ Solving the curves: $x_0^2\left(\frac{1}{a^2}-\frac{1}{{\alpha }^2}\right)=-y_0^2\left(\frac{1}{b^2}+\frac{1}{{\beta }^2}\right)\dots$ From (1) and (2): $\frac{1}{a^2}-\frac{1}{{\alpha }^2}=-\left(\frac{1}{b^2}+\frac{1}{{\beta }^2}\right)\frac{a^2{\alpha }^2}{b^2{\beta }^2}\Rightarrow a^2-b^2={\alpha }^2+{\beta }^2$