Question

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

• 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)

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

Given ellipse may be written as , $\frac{x^2}{9}+\frac{y^2}{4}=1.$

$\Rightarrow a^2=9,b^2=4,\therefore e=\sqrt{1-\frac{4}{9}}=\frac{\sqrt{5}}{3}$

$\Rightarrow F_1\equiv (-\sqrt{5},0),F_2\equiv (\sqrt{5},0)$

Let any point on the ellipse is, $P(3\cos \theta ,2\sin \theta )$
Thus area of triangle is $\Delta P{F}_1{F}_2=\frac{1}{2}|\left|\begin{array}{ccc}3\cos \theta & 2\sin \theta & 1\\ -\sqrt{5}& 0& 1\\ \sqrt{5}& 0& 1\end{array}\right||=2\sqrt{5}\sin \theta =\sqrt{10}$

$\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}\Rightarrow \theta =\pi /4,3\pi /4$
Corresponding options may be verified.