Question

Let $Q=\left(3,\sqrt{5}\right)$ and $R=\left(7,3\sqrt{5}\right)$ . A point $P$ in the $XY$ -plane varies in such a way that perimeter of $△PQR$ is $16$ . Then the maximum area of $△PQR$ is

Answer 1

Answer: 12

Now if we find distance between $Q$ and $R$ ,
$\Rightarrow QR=\sqrt{{\left(7-3\right)}^2+{\left(3\sqrt{5}-\sqrt{5}\right)}^2}=6$
And perimeter given $=16$
Therefore, $PQ+PR=10$ , which is greater than $6$ i.e. distance between given fixed points.
Therefore, point $P$ lies on the ellipse for which $Q$ and $R$ are foci and $PQ+PR=2a$
$\Rightarrow 2a=10\Rightarrow a=5$
$QR=2ae=6$
$\therefore a=5$ and $e=\frac{2ae}{2a}=\frac{6}{10}$
As we know $b^2=a^2-a^2{e}^2=25-9=16$
$\therefore b=4$
Maximum area of triangle $=\frac{1}{2}\left(2ae\right)\left(b\right)=12$ square units