Question

A point $P$ on the ellipse $4x^2+9y^2=36$ is such that $△P{F}_1{F}_2=\sqrt{10}$, where $F_1$ and $F_2$ are foci. Possible coordinates of $P$ are

• 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)

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

$\frac{x^2}{9}+\frac{y^2}{4}=1$
Let the coordinates of point $P$ are $(3\cos \theta ,2\sin \theta )$

$\therefore$ Area of the triangle $P{F}_1{F}_2$ is
$Area=\frac{1}{2}\left|\begin{array}{ccc}3cos\theta & 2sin\theta & 1\\ \sqrt{5}& 0& 1\\ -\sqrt{5}& 0& 1\end{array}\right|=\sqrt{10}$
$\Rightarrow sin\theta (2\sqrt{5)}=\pm \sqrt{10}$
$sin\theta =\pm \frac{\sqrt{2}}{2}=\pm \frac{1}{\sqrt{2}}$ and
$cos\theta =\pm \frac{1}{\sqrt{2}}$
There are 4 such points on the ellipse.
Options 'A' and 'D' are correct.