Question

Suppose $S$ and $S\prime$ 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 $PSS\prime$ , then the maximum value of $\Delta$ is double of

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

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 $S^{\prime }$ are $(3,0)$ and $(-3,0)$ respectively.
Let, $P\left(5cos\theta ,4sin\theta \right)$ be a variable point on the ellipse.
Then, $\Delta =areaof\Delta PS{S}^{\prime }=\frac{1}{2}\left|\begin{array}{ccc}3& 0& 1\\ -3& 0& 1\\ 5cos\theta & 4sin\theta & 1\end{array}\right|=12sin\theta$
Since the value of $sin\theta$ lies between $-1$ and $1$
So, the maximum value of area of $\Delta PS{S}^{\prime }$ is $12$ sq. units
Also, $a^8+b^4\ge 2a^4{b}^2(since,A.M\ge G.M)$
$\Rightarrow \frac{3\left(a^8+b^4\right)}{a^4{b}^2}\ge 6$