Question

Suppose $S$ and $S′$ are foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1.$ If $P$ is a variable point on the ellipse and if $\Delta $ is the area (in sq. units) of the triangle $PSS′$ , then the maximum value of $\Delta $ is double of

• A. Minimum value of $\frac{2 a^{8} + 2 b^{4}}{a^{4} b^{2}} \, \forall a, \, b\in R$
• B. Minimum value of $\frac{3 a^{8} + 3 b^{4}}{a^{4} b^{2}}$ $ \, \forall a, \, b\in R$
• C. $\frac{4 a^{8} + 4 b^{4}}{a^{4} b^{2}}$ $\forall a, \, b\in R$
• D. $\frac{6 a^{8} + 6 b^{4}}{a^{4} b^{2}}$ $\forall a, \, b\in R$

Answer 1

Answer: (B)

Given , $a^{2}=25$ and $b^{2}=16$
$\therefore \, e=\sqrt{1 - \frac{b^{2}}{a^{2}}}=\sqrt{1 - \frac{16}{25}}= \, \frac{3}{5}$
So, the coordinates of foci $S$ and $S^{′}$ are $\left(\right.3,0\left.\right)$ and $\left(\right.-3, \, 0\left.\right)$ respectively.
Let, $P\left(5 cos \theta , 4 sin ⁡ \theta \right)$ be a variable point on the ellipse.
Then, $\Delta =area \, of \, \Delta PSS^{′}=\frac{1}{2}\begin{vmatrix} 3 & 0 & 1 \\ -3 & 0 & 1 \\ 5cos \theta & 4sin⁡\theta & 1 \end{vmatrix}=12sin⁡\theta \, \, \, $
Since the value of $sin \theta \, $ lies between $-1$ and $1$
So, the maximum value of area of $\Delta PSS^{′}$ is $12$ sq. units
Also, $a^{8}+b^{4}\geq 2a^{4}b^{2}\left(\right.since,A.M\geq G.M\left.\right)$
$\Rightarrow \frac{3 \left(a^{8} + b^{4}\right)}{a^{4} b^{2}}\geq 6$