Question

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

• A. $(\pm \frac{3}{\sqrt{2}},\sqrt{2})$
• B. $(\frac{3}{2},2)$
• C. $(\frac{-3}{2},-2)$
• D. NONE

Answer 1

Answer: (A)

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

$4x^2+9y^2=36\Rightarrow \frac{x^2}{9}+\frac{y^2}{4}=1$

So, $a=3,b=2$

In an ellipse, Focal points, $F=(\pm \sqrt{a^2-b^2},0)=(\pm \sqrt{5},0)$

So, $F_1{F}_2=2\sqrt{5}$ which is the base of the triangle $P{F}_1{F}_2$

An arbitrry point P on ellipse is $(acos\theta ,b\sin \theta )$

So, $Heightof△P{F}_1{F}_2=b\sin \theta =2\sin \theta$

ie, $Area=\frac{1}{2}base\times height=\frac{1}{2}\times 2\sqrt{5}\times 2\sin \theta =2\sqrt{5}\sin \theta =\sqrt{10}$

$\Rightarrow \sin \theta =\frac{1}{\sqrt{2}}$

$\Rightarrow \cos \theta =\pm \frac{1}{\sqrt{2}}$

And $P=(a\cos \theta ,b\sin \theta )=(\pm \frac{3}{\sqrt{2}},\sqrt{2})$

So, Option$A$ has one of the anwers.