Question

Let $F_1$ and $F_2$ be the foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{17}=1$ and $M=|P_i{F}_1–P_i{F}_2|,i=1,2,3,4$ where $P_1,P_2,P_3,P_4$ are four points on the curve $4x^2–4xy+y^2–81=0$ such that either $M$ is greatest or least. If $S$ is set of distances between any $2$ different points of $P_i,$ then $S_{max}+S_{min}=$

Answer 1

$4x^2-4xy+y^2=81$
$\Rightarrow (2x-y{)}^2=81\Rightarrow 2x-y=\pm 9$
$\Rightarrow 2x-y-9=0$ or $2x-y+9=0$
$P_1,P_2,P_3,P_4$ lie on the pair of lines so that $|P_i{F}_1–P_i{F}_2|$ is either greatest or least.
$\therefore$ These points must be either collinear with $F_1,F_2$ or lie on the perpendicular bisector of $F_1{F}_2.$
Quadrilateral $P_1{P}_2{P}_3{P}_4$ is a rhombus as shown in the figure.

$S=\overline{P_i{P}_j}(i,j=1,2,3,4\&i\ne j)$
$S_{max}=18,S_{min}=9$
$\therefore S_{max}+S_{min}=27$