Question

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

Answer 1

Given, Equation of the ellipse,

$\frac{x^2}{16}+\frac{y^2}{9}=1$

$\therefore$ Chord of contact, $\frac{x{x}^1}{16}+\frac{y{y}^1}{9}=1\Rightarrow \frac{x\times 0}{16}+\frac{y\times 5}{9}=1\Rightarrow y=\frac{9}{5}$

Putting $y=\frac{9}{5}$ in the above equation of the ellipse,

$\frac{x^2}{16}+\frac{{\left(\frac{9}{5}\right)}^2}{9}=1\Rightarrow \frac{x^2}{16}=\frac{16}{25}\Rightarrow x^2=\frac{16\times 16}{25}\Rightarrow x=\pm \frac{16}{5}$

So, $A=(\frac{16}{5},\frac{9}{5}),B=(-\frac{16}{5},\frac{9}{5}),P=(0,5)$

Since, $△PAB$ is isosceles with base $AB$

$\therefore Height=(5-\frac{9}{5})units=\frac{16}{5}units$

$\therefore$ Area of $△PAB=\frac{1}{2}\times Base\times Height=\frac{1}{2}\times (\frac{16}{5}+\frac{16}{5})\times \frac{16}{5}=\frac{256}{25}squareunits.$