Question

A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$ intersects the major and minor axes at point $A$ and $B$ respectively. If $C$ is the centre of the ellipse, then the area of the triangle $ABC$ (in sq. unit) is

Answer 1

One of the tangent with slope $m$ to the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$ is:
$y=mx+\sqrt{18m^2+32}$
Given, $m=-\frac{4}{3}$
$\Rightarrow y=-\frac{4}{3}x+8$
This line intersect coordinate axis at $A(0,8),B(6,0)$ and $C$ is $(0,0)$
Then, area of triangle $ABC$
$=\frac{1}{2}\left(6\right)\left(8\right)=24\text{sq. units}$


Alternate Solution:
Equation of tangents with slope $m=-\frac{4}{3}$ to the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1$ is :
$y=mx\pm \sqrt{18m^2+32}$
$\Rightarrow y=-\frac{4}{3}x\pm 8$
This line intersect coordinate axis at $A(0,\pm 8),B(\pm 6,0)$ and $C$ is $(0,0).$
If we consider first quadrant, then coordinates are $A(0,8),B(6,0)$ and $C(0,0).$
Then, area of triangle $ABC$
$=\frac{1}{2}\left(6\right)\left(8\right)=24$ sq. unit.
If we consider third quadrant, then coordinates are $A(0,-8),B(-6,0)$ and $C(0,0).$
Then, area of triangle $ABC$
$=\frac{1}{2}\left(6\right)\left(8\right)=24$ sq. unit.