Q90 / 150 · MHT CET 2009 · Mathematics
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 $\Delta A O B$, where $O$ is the origin.
- A12 sq unit
- B$8 \mathrm{sq}$ unit
- Correct answerC24 sq unit
- D32 sq unit
Solution
Equation of tangent with slope $-\frac{3}{4}$ is
$$
y=-\frac{3}{4} x+c
$$
According to condition of tengency
$\therefore$
$$
\begin{aligned}
c &=\sqrt{32 \times\left(\frac{-3}{4}\right)^{2}+18} \\
&=\sqrt{18+18} \\
&=6 \\
y &=-\frac{3}{4} x+6
\end{aligned}
$$
$$
\Rightarrow 4 y+3 x=24
$$
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
Hence, the correct answer is option: 'C'.