Question Number 62753 by peter frank last updated on 24/Jun/19
$${The}\:{normal}\:{at}\:{the}\:{point} \\ $$$${P}\left(\mathrm{4cos}\:\theta,\mathrm{3sin}\:\theta\right)\:{on}\:{the} \\ $$$${ellipse}\:\frac{{x}^{\mathrm{2}} }{\mathrm{16}}\:+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1}\:{meets} \\ $$$${the}\:{x}−{axis}\:{and}\:{y}−{axis} \\ $$$${at}\:{A}\:{and}\:{B}\:{respectively} \\ $$$${show}\:{that}\:{locus}\:{of}\:{the} \\ $$$${mid}−{point}\:{of}\:{AB}\:{is}\:{an} \\ $$$${ellipse}\:{with}\:{the}\:{same} \\ $$$${eccentricity}\:{as}\:{given} \\ $$$${ellipse}. \\ $$
Answered by Hope last updated on 25/Jun/19
$${eqn}\:{normal}\:\left({y}−\mathrm{3}{sin}\theta\right)=\left(\frac{−{dx}}{{dy}}\right)_{\left(\mathrm{4}{cos}\theta,\mathrm{3}{sin}\theta\right)} \:\left({x}−\mathrm{4}{cos}\theta\right) \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{16}}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1} \\ $$$$\frac{\mathrm{2}{x}×\frac{{dx}}{{dy}}}{\mathrm{16}}+\frac{\mathrm{2}{y}}{\mathrm{9}}=\mathrm{0} \\ $$$$\frac{{x}×\frac{{dx}}{{dy}}}{\mathrm{8}}=\frac{−\mathrm{2}{y}}{\mathrm{9}}\rightarrow\frac{{dx}}{{dy}}=\frac{−\mathrm{16}{y}}{\mathrm{9}{x}}=\frac{−\mathrm{16}×\mathrm{3}{sin}\theta}{\mathrm{9}×\mathrm{4}{cos}\theta}=\frac{−\mathrm{4}}{\mathrm{3}}{tan}\theta \\ $$$$\left({y}−\mathrm{3}{sin}\theta\right)=\frac{\mathrm{4}{tan}\theta}{\mathrm{3}}\left({x}−\mathrm{4}{cos}\theta\right) \\ $$$${put}\:{x}=\mathrm{0} \\ $$$${y}−\mathrm{3}{sin}\theta=\frac{\mathrm{4}{sin}\theta}{\mathrm{3}{cos}\theta}×−\mathrm{4}{cos}\theta \\ $$$${y}=\mathrm{3}{sin}\theta−\frac{\mathrm{16}{sin}\theta}{\mathrm{3}}=\frac{−\mathrm{7}{sin}\theta}{\mathrm{3}} \\ $$$${B}\left(\mathrm{0},\frac{−\mathrm{7}{sin}\theta}{\mathrm{3}}\right) \\ $$$${put}\:{y}=\mathrm{0} \\ $$$$−\mathrm{3}{sin}\theta=\frac{\mathrm{4}{sin}\theta}{\mathrm{3}{cos}\theta}\left({x}−\mathrm{4}{cos}\theta\right) \\ $$$${x}=\frac{−\mathrm{9}{sin}\theta{cos}\theta}{\mathrm{4}{sin}\theta}+\mathrm{4}{cos}\theta \\ $$$${x}=\frac{−\mathrm{9}{cos}\theta+\mathrm{16}{cos}\theta}{\mathrm{4}}=\frac{\mathrm{7}{cos}\theta}{\mathrm{4}}\boldsymbol{{A}}\left(\frac{\mathrm{7}{cos}\theta}{\mathrm{4}},\mathrm{0}\right) \\ $$$${Mid}\:{point}\:{of}\:{A}\boldsymbol{{B}}=\left(\frac{\mathrm{7}{cos}\theta}{\mathrm{8}},\frac{−\mathrm{7}{sin}\theta}{\mathrm{6}}\right) \\ $$$${locus}\:\alpha=\frac{\mathrm{7}{cos}\theta}{\mathrm{8}}\:\:\:\beta=\frac{−\mathrm{7}{sin}\theta}{\mathrm{6}} \\ $$$$\left(\frac{\mathrm{8}\alpha}{\mathrm{7}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{6}\beta}{−\mathrm{7}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${locus}\:\:\frac{{x}^{\mathrm{2}} }{\left(\frac{\mathrm{7}}{\mathrm{8}}\right)^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\left(\frac{\mathrm{7}}{\mathrm{6}}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$${eccenrixity}=\frac{\sqrt{\left(\frac{\mathrm{7}}{\mathrm{6}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{7}}{\mathrm{8}}\right)^{\mathrm{2}} }}{\frac{\mathrm{7}}{\mathrm{6}}}=\frac{\mathrm{6}}{\mathrm{7}}×\mathrm{7}×\sqrt{\frac{\mathrm{64}−\mathrm{36}}{\mathrm{8}^{\mathrm{2}} ×\mathrm{6}^{\mathrm{2}} }} \\ $$$${e}_{\mathrm{2}} =\frac{\mathrm{6}}{\mathrm{7}}×\mathrm{7}×\frac{\mathrm{1}}{\mathrm{8}×\mathrm{6}}×\mathrm{2}\sqrt{\mathrm{7}}\:=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{16}}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1}\:\:{e}_{\mathrm{1}} =\frac{\sqrt{\mathrm{16}−\mathrm{9}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{4}} \\ $$$${so}\:{e}_{\mathrm{1}} ={e}_{\mathrm{2}} =\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}{proved} \\ $$$$ \\ $$$$ \\ $$