Question

A tangent having slope $-\frac{4}{3}$ to the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$, intersects the axis of $x$ and $y$ in points $A\&B$ respectively. If $O$ is the origin, find the area of triangle $OAB$.

Answer 1

Ellipse:$\frac{x^2}{18}+\frac{y^2}{32}=1$ $(b>a)$

slope of tangent$=-\frac{4}{3}$, let it be $y=-\frac{4}{3}+c$

$c=\pm \sqrt{b^2{m}^2+a^2}$

$c=\pm \sqrt{32\frac{\left(16\right)}{9}+18}$

$c=\pm 8$

$T:y=-\frac{4}{3}x\pm 8$

intercepts:$x=0,y=8$

$y=0,x=6$

$Ar.△OAB=\frac{1}{2}\left(OA\right)\left(OB\right)=\frac{1}{2}\left(8\right)\left(6\right)=24$