Question

If 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 in points $A$ and $B$ respectively, then the area of $△OAB$ is equal to $(O$ is the centre of the ellipse)

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

Answer 1

Answer: (D)

$24sq.units$

Let $P(x_1,y_1)$ be a point on the ellipse.
$\frac{x^2}{18}+\frac{y^2}{32}=1$
$\Rightarrow \frac{x_1^2}{18}+\frac{y_1^2}{32}=1....\left(i\right)$
The equation of the tangent at $(x_1,y_1)$ is $\frac{x{x}_1}{18}+\frac{y{y}_1}{32}=1$. This meets the axes at $A(\frac{18}{x_1},0)$ and $B(0,\frac{32}{y_1})$. It is given that slope of the tangent at $(x_1,y_1)$ is $-\frac{4}{3}$
So, $-\frac{x_1}{18}\cdot \frac{32}{y_1}=-\frac{4}{3}$
$\Rightarrow \frac{x_1}{y_1}=\frac{3}{4}$
$\Rightarrow \frac{x_1}{3}=\frac{y_1}{4}=K$ (say)
$\therefore x_1=3K$ and $y_1=4K$
Putting $x_1,y_1$ in (i), we get
$K^2=1$
$\therefore$ Area of $△OAB=\frac{1}{2}OA.OB$
$=\frac{1}{2}\cdot \frac{18}{x_1}\cdot \frac{32}{y_1}=\frac{1}{2}\frac{\left(18\right)\left(32\right)}{\left(3K\right)\left(4K\right)}=\frac{24}{K^2}$
$=24squnits(\because K^2=1)$

Hence, option D is correct.