Question

Consider an ellipse $E: \frac{(3 x-4 y+5)^2}{100}+\frac{(4 x+3 y-10)^2}{50}=1 . M$ 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 $m \leq p_1+p_2 \leq M$, then $P$ lies in the region whose area is $\Delta$. Find the sum of the digits in $\Delta$

Answer 1

$\frac{\left(\frac{3 x-4 y+5}{5}\right)^2}{4}+\frac{\left(\frac{4 x+3 y-10}{5}\right)^2}{2}=1$
$\therefore m =4, m =2 \sqrt{2} $
$\text { Major axis : } 4 x +3 y -10=0$
$\text { minor axis : } 3 x -4 y +5=0$
$m \leq p _1+ p _2 \leq M$
$2 \sqrt{2} \leq\left|\frac{4 x +3 y -10}{5}\right|+\left|\frac{3 x -4 y +5}{5}\right| \leq 4 $
$2 \sqrt{2} \leq| x |+| y | \leq 4 $
$\text { Required area }=32-16=16$