Question Number 47659 by behi83417@gmail.com last updated on 12/Nov/18
Answered by ajfour last updated on 13/Nov/18
$${Eq}.\:{of}\:{tangent}: \\ $$$$\frac{{xx}_{\mathrm{1}} }{{a}^{\mathrm{2}} }+\frac{{yy}_{\mathrm{1}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${A}\left(\frac{{a}^{\mathrm{2}} }{{x}_{\mathrm{1}} },\mathrm{0}\right)\:\:\:;\:{B}\left(\mathrm{0},\frac{{b}^{\mathrm{2}} }{{y}_{\mathrm{1}} }\right) \\ $$$${Area}\:\bigtriangleup{AOB}\:=\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{\mathrm{2}\mid{x}_{\mathrm{1}} {y}_{\mathrm{1}} \mid} \\ $$$$\:\:\:{with}\:\:\frac{{x}_{\mathrm{1}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}_{\mathrm{1}} ^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${So}\:{Area}\:\bigtriangleup{AOB}\:=\:\frac{{a}^{\mathrm{2}} {b}}{\mathrm{2}\mid{x}_{\mathrm{1}} \mid\sqrt{\mathrm{1}−\frac{{x}_{\mathrm{1}} ^{\mathrm{2}} }{{a}^{\mathrm{2}} }}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{{a}^{\mathrm{3}} {b}}{\mathrm{2}\sqrt{{x}_{\mathrm{1}} ^{\mathrm{2}} \left({a}^{\mathrm{2}} −{x}_{\mathrm{1}} ^{\mathrm{2}} \right)}}\: \\ $$$${with}\:\:\:{x}_{\mathrm{1}} \in\left(−{a},{a}\right)\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18
$${tangent}\:{at}\:\left({acos}\theta,{bsin}\theta\right) \\ $$$$\frac{{x}×{acos}\theta}{{a}^{\mathrm{2}} }+\frac{{y}×{bsin}\theta}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}}{\frac{{a}}{{cos}\theta}}+\frac{{y}}{\frac{{b}}{{sin}\theta}}=\mathrm{1} \\ $$$${A}\left(\frac{{a}}{{cos}\theta},\mathrm{0}\right)\:\:\:{B}\left(\mathrm{0},\frac{{b}}{{sin}\theta}\right) \\ $$$${area}\:{triangle}\:{OAB}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{a}}{{cos}\theta}×\frac{{b}}{{sin}\theta}=\frac{{ab}}{{sin}\mathrm{2}\theta} \\ $$$$\:\:\: \\ $$