Question

A tangent to the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$, having slope $-\frac{4}{3}$ meets the major and minor axes at $A$ and $B$. If $O$ be the origin, then the area of $△OAB$ is

Answer 1

General equation of tangent for ellipse with eccentric angle $\theta$ is

$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

Given: Slope $=-\frac{4}{3}$

$\Rightarrow \frac{-b}{a\tan \theta }=\frac{-4}{3}$

$\Rightarrow \frac{\sqrt{32}}{\sqrt{18}\tan \theta }=\frac{4}{3}$

$\Rightarrow \tan \theta =1$

$\Rightarrow \theta ={45}^o$

Tangent will meet minor axis at $(\frac{a}{cos\theta },0)=(\frac{\sqrt{18}}{cos45},0)=(6,0)$

Tangent will meet major axis at $(0,8)$

Area of triangle $AOB$ is $\frac{1}{2}\times 8\times 6=24$

$\therefore$ Area is $24$ units