Question

A rectangle $ABCD$ has area $200.$Sq.units and an ellipse with area $200\pi$ Sq.units having foci at $B$ and $D$ passes through $A$ and $C$ . If the perimeter of the rectangle is $P$ units then the value of $\frac{P}{20}$ is

Answer 1

Let side of rectangle be $p$ and $q$
Area of rectangle$=pq=200...\left(1\right)$

Area of ellipse $=\pi ab=200\pi$
$\therefore ab=200...\left(2\right)$
we have to find the perimeter of rectangle $=2(p+q)$
From triangle $ABD$
Distance $BD=\sqrt{p^2+q^2}=$ distance between foci
or
$p^2+q^2=4a^2{e}^2$
$(p+q{)}^2-2pq=4(a^2-b^2)...(3)$
Also from the definition of ellipse, sum of focal lengths is $2a$,
Then
$AB+AD=p+q=2a...\left(4\right)$
putting value of $(p+q)$ in equation $\left(3\right)$ and $\left(4\right)$
we have $(2a{)}^2-2pq=4a^2-4b^2$ (using equation $\left(1\right)$)
$\Rightarrow 4a^2-2\times 200=4(a^2-b^2)$
$\Rightarrow a^2-100=a^2-b^2$
$\Rightarrow b=10$
from equation $\left(2\right),ab=200\Rightarrow a=20$
Since $p+q=2a$ (from equation $\left(4\right)$)
$\therefore$ perimeter$=2(p+q)=4a=4\times 20=80$
and
$\frac{P}{4}=\frac{80}{20}=4$