Question

If a tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{x^{2}}{18}+\frac{y^{2}}{34}=1$ intersects the major and minor axes in points $A$ and $B$ respectively, then the area of $\Delta OAB$ is equal to (O is the centre of the ellipse)

• A. 12 sq units
• B. 48 sq units
• C. 64 sq units
• D. 24 sq units

Answer 1

The Correct Option is D

Let $P\left(x_{1}, y_{1}\right)$ be a point on the ellipse.
$ \frac{x^{2}}{18}+\frac{y^{2}}{32}=1 $
$\Rightarrow \frac{x_{1}^{2}}{18}+\frac{y_{1}^{2}}{32}=1$
The equation of the tangent at $\left(x_{1}, y_{1}\right)$ is
$\frac{x x_{1}}{18}+\frac{y y_{1}}{32}=1$.
This meets the axes at $A\left(\frac{18}{x_{1}}, 0\right)$ and
$B\left(0, \frac{32}{y_{1}}\right)$ It is given that slope of the tangent at
$\left(x_{1}, y_{1}\right)$ is $-\frac{3}{4}$
Hence, $-\frac{x_{1}}{18} \cdot \frac{32}{y_{1}}=-\frac{4}{3}$
$\Rightarrow \frac{x_{1}}{y_{1}}=\frac{3}{4}$
$\Rightarrow \frac{x_{1}}{3}=\frac{y_{1}}{4}=K $(say)
$ \therefore x_{1}=3 K $
$ y_{1}=4 K$
Putting $x_{1}, y_{1}$ in (i), we get
$K^{2}=1$
Now, are of $\Delta O A B=\frac{1}{2} O A \cdot O B$
$=\frac{1}{2} \cdot \frac{18}{x_{1}} \cdot \frac{32}{y_{1}}$
$=\frac{1}{2} \frac{(18)(32)}{\left(x_{1} y_{1}\right)}$
$=\frac{1}{2} \frac{(18)(32)}{(3 K)(4 K)}$
$=\frac{24}{K^{2}}$
$=24$ sq units $\left(\because K^{2}=1\right)$