Question

Tangent to the ellipse $\frac{x^2}{32}+\frac{y^2}{18}=1$ having slope $\frac{-3}{4}$ meet the coordinate axis at $A$ and $B$. Then, the area of $△AOB$, where $O$ is the origin, is

• A. $12$ sq.units
• B. $8$ sq.units
• C. $24$ sq. units
• D. $32$ sq. units

Answer 1

The correct option is C $24$ sq. units

Equation of tangent to ellipse in slope form is

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

So equation of tangent to $\frac{x^2}{32}+\frac{y^2}{18}=1$

$y=-\frac{3}{4}x\pm \sqrt{32\times \frac{9}{16}+18}y=-\frac{3}{4}x\pm \sqrt{36}y=-\frac{3}{4}x\pm 64y+3x=\pm 24$

Put $x=0\Rightarrow y=\pm 6$

Put $y=0\Rightarrow x=\pm 8$

Area $△AOB$ $=\frac{1}{2}|\pm 6||\pm 8|=24$

So option $C$ is correct.