Question

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

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

Answer 1

The correct option is C $24$sq unit

Equation of tangent with slope$-\frac{3}{4}$ is $y=-\frac{3}{4}x+c$
According to condition of tengency $c=\sqrt{32\times {\left(\frac{-3}{4}\right)}^2+18}$
$=\sqrt{18+18}$
$=6$
$\therefore y=-\frac{3}{4}x+6$
$\Rightarrow 4y+3x=24$
It meets the coordinate axes in A and B.
$\therefore A\equiv (8,0)$ and $B\equiv (0,6)$
Required area $=\frac{1}{2}\times 8\times 6=24$sq unit.