Sep 2019 CSAT mock test for Sciences, #21

The value of $\rm \overline{PA}+\overline{PB}$ is determined by the size of the ellipse that passes through point $\rm P$ whose foci are $\rm A$ and $\rm B.$

Hence the rectangle touches the ellipse $\dfrac{x^2}{40}+\dfrac{y^2}{36}=1$ at ${\rm X}(0,~6),$ and since $\rm \overline{PA}+\overline{PB}=2\sqrt{10},$ the rectangle touches the ellipse $\dfrac{x^2}{10}+\dfrac{y^2}{6}=1$ at ${\rm Y}\left(\dfrac{5}{2},~\dfrac{3}{2}\right).$

Then, note that the tangent line of the ellipse $\dfrac{x^2}{10}+\dfrac{y^2}{6}=1$ is $y=-x+4.$ Now we know that the other side of the rectangle should be on $y=-x+6.$ (A line with the same slope, passing through $X,$ that is.)

To find the intersection between $y=-x+6$ and $\dfrac{x^2}{40}+\dfrac{y^2}{36}=1,$ we solve $\dfrac{x^2}{40}+\dfrac{(-x+6)^2}{36}=1 \\ \text{} \\ 19x^2-120x=0 \\ \text{} \\ x=\dfrac{120}{19}.$

Hence ${\rm Z}\left(\dfrac{120}{19},\,-\dfrac{6}{19}\right).$ The slope of the ellipse at $\rm Z$ is $18,$ and hence the rectangle is completely inside the ellipse, and this rectangle is the biggest it can ever get.

$\therefore~\dfrac{q}{p}=\dfrac{120}{19}\sqrt{2}\times\sqrt{2}=\dfrac{240}{19},\quad p+q=19+240=259.$