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Question Number 47659 by behi83417@gmail.com last updated on 12/Nov/18

Answered by ajfour last updated on 13/Nov/18

$$Eq.oftangent:$$$$\frac{{xx}_1}{a^2}+\frac{{yy}_1}{b^2}=1$$$$A(\frac{a^2}{x_1},0);B(0,\frac{b^2}{y_1})$$$$Area△AOB=\frac{a^2{b}^2}{2\mid x_1{y}_1\mid }$$$$with\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1$$$$SoArea△AOB=\frac{a^{2}b}{2\mid x_1\mid \sqrt{1-\frac{x_1^2}{a^2}}}$$$$=\frac{a^{3}b}{2\sqrt{x_1^2(a^2-x_1^2)}}$$$$with{x}_1\in (-a,a).$$

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Nov/18

$$tangentat(acos\theta ,bsin\theta )$$$$\frac{x\times acos\theta }{a^2}+\frac{y\times bsin\theta }{b^2}=1$$$$\frac{x}{\frac{a}{cos\theta }}+\frac{y}{\frac{b}{sin\theta }}=1$$$$A(\frac{a}{cos\theta },0)B(0,\frac{b}{sin\theta })$$$$areatriangleOAB=\frac{1}{2}\times \frac{a}{cos\theta }\times \frac{b}{sin\theta }=\frac{ab}{sin2\theta }$$$$$$

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