Question Number 51822 by peter frank last updated on 30/Dec/18
$${If}\:\:{p}\:{and}\:{q}\:\:{are}\:{the}\:{length} \\ $$$${of}\:{perpendicular}\:{from} \\ $$$${the}\:{origin}\:{to}\:{the}\:{lines} \\ $$$${x}\mathrm{cos}\:\theta−{y}\mathrm{sin}\:\:\theta={kcos}\mathrm{2}\theta \\ $$$${and}\:{x}\mathrm{sec}\:\theta+{y}\mathrm{cosec}\:\theta={k} \\ $$$${respectively} \\ $$$${prove}\:{that} \\ $$$${p}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{2}} ={k}^{\mathrm{2}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Dec/18
$${p}=\mid\frac{\mathrm{0}+\mathrm{0}−{k}}{\:\sqrt{{cos}^{\mathrm{2}} \theta+{sin}^{\mathrm{2}} \theta}}\mid={k} \\ $$$${q}=\mid\frac{−{k}}{\:\sqrt{{sec}^{\mathrm{2}} \theta+{cosec}^{\mathrm{2}} \theta}}\mid=\frac{{k}}{\:\sqrt{\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}+\frac{\mathrm{1}}{{sin}^{\mathrm{2}} \theta}}}={ksin}\theta{cos}\theta \\ $$$${pls}\:{check}\:{the}\:{question}… \\ $$
Commented by peter frank last updated on 30/Dec/18
$${now}\:{fixed} \\ $$
Commented by peter frank last updated on 30/Dec/18
$${p}^{\mathrm{2}} ={k}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\mathrm{2}\theta….\left({i}\right) \\ $$$${q}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}{k}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \mathrm{2}\theta….\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right) \\ $$$${p}^{\mathrm{2}} +\mathrm{4}{q}^{\mathrm{2}} ={k}^{\mathrm{2}} \\ $$$${thank}\:{you}\:{sir}\:{for}\:{your}\:{method}\:…..{finaly}\:{i}\:{reach} \\ $$$${destination} \\ $$