Question Number 226338 by Spillover last updated on 25/Nov/25
Answered by Spillover last updated on 06/Dec/25
$${at}\:{point}\:{P}\left({a}\mathrm{cos}\:\theta+{b}\mathrm{sin}\:\theta\right){aa} \\ $$$$ \\ $$$${at}\:{point}\:{Q}\left({a}\mathrm{cos}\:\left(\theta+\frac{\pi}{\mathrm{2}}\right)+{i}\mathrm{sin}\left(\theta+\frac{\pi}{\mathrm{2}}\right)\right. \\ $$$${Q}=\left({a}\mathrm{sin}\:\theta,{b}\mathrm{cos}\:\theta\right. \\ $$$${from}\:{equation}\:{of}\:{normal} \\ $$$$\frac{{a}^{\mathrm{2}} {x}}{{a}\mathrm{cos}\:\:\theta}−\frac{{by}}{\mathrm{sin}\:\theta}={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$\theta=\theta \\ $$$$\theta=\theta+\frac{\pi}{\mathrm{2}} \\ $$$$\frac{{ax}}{{a}\mathrm{cos}\:\left(\:\theta+\frac{\pi}{\mathrm{2}}\right)}−\frac{{by}}{\mathrm{sin}\:\left(\theta+\frac{\pi}{\mathrm{2}}\right)}={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$$−\frac{{ax}}{\mathrm{sin}\:\theta}−\frac{{by}}{\mathrm{cos}\:\theta}=\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right) \\ $$$$\frac{{ax}}{\mathrm{sin}\:\theta}+\frac{{by}}{\mathrm{cos}\:\theta}=\left({b}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$…… \\ $$$$\: \\ $$