Question

If point $P$ lies on the ellipse $4x^2+9y^2=36$ such that $\Delta P{F}_1{F}_2=\sqrt{10}$ , where $F_1$ and $F_2$ are foci. Then $P$ can 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)

The correct options are
A $(\frac{3}{\sqrt{2}},\sqrt{2})$
D $(\frac{-3}{\sqrt{2}},-\sqrt{2})$

$\frac{x^2}{9}+\frac{y^2}{4}=1$
Any point on ellipse can be written as $(3\cos \theta ,2\sin \theta )$
Area of triangle is $=\frac{1}{2}|2\sin \theta |F_1{F}_2=|\sin \theta |2ae$
$e^2=\frac{a^2-b^2}{a^2}=\frac{9-4}{9}$
$e=\frac{\sqrt{5}}{3}$
Area$=|\sin \theta |2\times 3\times \frac{\sqrt{5}}{3}=|\sin \theta |\times 2\sqrt{5}=\sqrt{10}$
$⟹\sin \theta =\pm \frac{1}{\sqrt{2}}$
So P can be$(\frac{3}{\sqrt{2}},\sqrt{2})$ or $(-\frac{3}{\sqrt{2}},-\sqrt{2})$
Hence, options 'A' and 'D' are correct.