JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Ellipse

For the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1$, we have $a^2=9,b^2=8$, so $c=1$ and foci are $F_1(-1,0)$ and $F_2(1,0)$. The parabola with vertex at origin and focus at $F_2(1,0)$ is $y^2=4x$. Solving with the ellipse: $\frac{x^2}{9}+\frac{4x}{8}=1$ gives $x=\frac{3}{2}$, so $M=(\frac{3}{2},\sqrt{6})$ and $N=(\frac{3}{2},-\sqrt{6})$. Since $MN$ is vertical, one altitude is the $x$-axis ($y=0$). The other altitude from $M$ perpendicular to $F_{1}N$ has slope $\frac{5}{2\sqrt{6}}$. The equation is $(y-\sqrt{6})=\frac{5}{2\sqrt{6}}(x-\frac{3}{2})$. Setting $y=0$: $x=-\frac{9}{10}$. Thus the orthocentre is $(-\frac{9}{10},0)$.