Question

The point $P$ on the ellipse $4x^2+9y^2=36$ is such that the area of the $△P{F}_1{F}_2=\sqrt{10}\text{sq. units}$, where $F_1,F_2$ are foci. Then the coordinates of $P$ will be

• A. $(\frac{3}{\sqrt{2}},\sqrt{2})$
• B. $(\frac{3}{2},2)$
• C. $(\frac{-3}{2},-2)$
  
• D. $(\frac{-3}{\sqrt{2}},-\sqrt{2})$

Answer 1

Answer: (A), (D)

$(\frac{-3}{\sqrt{2}},-\sqrt{2})$

Given, equation of ellipse
$\frac{x^2}{9}+\frac{y^2}{4}=1\therefore e=\sqrt{1-\frac{4}{9}}\Rightarrow e=\frac{\sqrt{5}}{3}$

Coordinates of foci $F_1$ and $F_2$ are
$(\pm \sqrt{5},0)$
Let $P\equiv (h,k)$
Now, the area of
$△P{F}_1{F}_2=\sqrt{10}\Rightarrow \frac{1}{2}\times 2\sqrt{5}\times |k|=\sqrt{10}\Rightarrow |k|=\sqrt{2}\Rightarrow k=\sqrt{2}\text{or}-\sqrt{2}\Rightarrow y=\sqrt{2}\text{or}-\sqrt{2}$

Putting value of $y$ in equation of ellipse
$\Rightarrow 4x^2+9(\sqrt{2}{)}^2=36\Rightarrow x^2=\frac{9}{2}\Rightarrow x=\frac{3}{\sqrt{2}},-\frac{3}{\sqrt{2}}$

Hence, possible points are
$(\frac{3}{\sqrt{2}},\sqrt{2}),(\frac{3}{\sqrt{2}},-\sqrt{2}),(-\frac{3}{\sqrt{2}},\sqrt{2}),(-\frac{3}{\sqrt{2}},-\sqrt{2})$