Rectangle $ABCD$has area 200. An ellipse with area $200\pi$passes through $A$and $C$ and has foci at $B$ and $D$. If the perimeter of the rectangle is $P$, then the value of $P/20$is ________

Let the sides of rectangle be $p\text{ and }q$

$\text{Area of rectangle}=pq=\mathrm{200.....}\left(i\right)$

$\text{Area of ellipse}=\pi ab=200\pi$

$\therefore ab=\mathrm{200....}\left(ii\right)$

$\text{Perimeter of rectangle}=2(p+q)$

From triangle $ABD$

$\text{Distance }BD=\sqrt{p^2+q^2}=\text{Distance between foci}$

$or{p}^2+q^2=4a^2{e}^2$

$or{(p+q)}^2-2pq=4(a^2-b^2)\mathrm{....}\left(iii\right)$

Also, from definition of ellipse, the sum of focal length is $2a$ Then, $AB+AD=p+q=2a\mathrm{....}\left(iv\right)$

Putting the value of $(p+q)$ in $\left(iii\right)$ from $\left(iv\right),$ we have

${\left(2a\right)}^2-2pq=4a^2-4b^2$

$or4a^2-2\times 200=4(a^2-b^2)$

$or{a}^2-100=a^2-b^2$

$orb=10$

From $\left(ii\right),$

$ab=200ora=20$

Since $p+q=2a$

$\text{Perimeter}=2(p+q)=4a=4\times 20=80$