Question

If Tangents $PA$ and $PB$ are drawn to ellipse
$\frac{x^2}{16}+\frac{y^2}{9}=1$ from a point $P(0,5),$ then area of triangle $PAB$ is equal to

• A. $\frac{16}{5}$ sq. unit
• B. $\frac{256}{25}$ sq. unit
• C. $\frac{32}{5}$ sq. unit
• D. $\frac{1024}{25}$ sq. unit

Answer 1

The correct option is B $\frac{256}{25}$ sq. unit

Equation of chord of contact of $(0,5)$ with respect to ellipse $S$ is $S_1=0$


$\Rightarrow \frac{x\left(0\right)}{16}+\frac{y\left(5\right)}{9}=1\Rightarrow y=\frac{9}{5}\dots \left(1\right)$
Now, solving $\left(1\right)$ and $\frac{x^2}{16}+\frac{y^2}{9}=1$
$A(\frac{-16}{5},\frac{9}{5});B(\frac{16}{5},\frac{9}{5})$
$\therefore$ Required Area of $△PAB$
$=2[\frac{1}{2}\times \frac{16}{5}\times (5-\frac{9}{5})]$
$=\frac{256}{25}$ sq. unit