Question

$P$ is a variable point on the ellipse $9x^2+16y^2=144$ with foci $S$ and $S^1$. If $K$ is the area of the triangle $S{S}^{1}P$, then the maximum value of $K$ is

• A. $7\sqrt{3}$
• B. $3\sqrt{5}$
• C. $7\sqrt{5}$
• D. $3\sqrt{7}$

Answer 1

The correct option is D $3\sqrt{7}$

Given ellipse may be written as , $\frac{x^2}{16}+\frac{y^2}{9}=1$
$\Rightarrow a^2=16,b^2=9,\therefore e=\sqrt{1-\frac{9}{16}}=\frac{\sqrt{7}}{4}$
$\Rightarrow S\equiv (-\sqrt{7},0),S^1\equiv (\sqrt{7},0)$
Let any point on the ellipse is, $P(4\cos \theta ,3\sin \theta )$
Thus area of triangle is $K=\frac{1}{2}\left|\left|\begin{array}{ccc}4\cos \theta & 3\sin \theta & 1\\ -\sqrt{7}& 0& 1\\ \sqrt{7}& 0& 1\end{array}\right|\right|=3\sqrt{7}\sin \theta$
We know maximum value of $\sin \theta$ is 1.
Hence, $K_{max}=3\sqrt{7}$