Question

Consider an ellipse $E:\frac{(3x-4y+5{)}^2}{100}+\frac{(4x+3y-10{)}^2}{50}=1.M$ and $m$ are lengths of major and minor axes respectively. If $p_1$ and $p_2$ are perpendicular distances on major and minor axes from a point $P$ in the $x-y$ plane such that $m\le p_1+p_2\le M$, then $P$ lies in the region whose area is $\Delta$. Find the sum of the digits in $\Delta$

Answer 1

Answer: 7

$\frac{{\left(\frac{3x-4y+5}{5}\right)}^2}{4}+\frac{{\left(\frac{4x+3y-10}{5}\right)}^2}{2}=1$
$\therefore m=4,m=2\sqrt{2}$
$\text{ Major axis : }4x+3y-10=0$
$\text{ minor axis : }3x-4y+5=0$
$m\le p_1+p_2\le M$
$2\sqrt{2}\le \left|\frac{4x+3y-10}{5}\right|+\left|\frac{3x-4y+5}{5}\right|\le 4$
$2\sqrt{2}\le |x|+|y|\le 4$
$\text{ Required area }=32-16=16$