Question

Let $F_1(x_1,0)$ and $F_2(x_2,0),$ where $x_1<0,x_2>0,$ be the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{8}=1.$ A parabola having vertex at the origin and focus at $F_1$ intersects the ellipse at the point $M$ in the second quadrant and at point $N$ in the third quadrant. If the distance between $M$ and $N$ is $d$, then the value of $d^2$ is

Answer 1

Given ellipse : $\frac{x^2}{9}+\frac{y^2}{8}=1\cdots \left(i\right)$
$a=3,b=2\sqrt{2}\Rightarrow e=\sqrt{1-\frac{b^2}{a^2}}\Rightarrow e=\sqrt{\frac{9-8}{9}}=\frac{1}{3}$

Foci $=(\pm ae,0)=(\pm 1,0)$
$\therefore F_1=(-1,0),F_2=(1,0)$

Now, the equation of the parabola having vertex at the origin, and focus at $F_1(-1,0)$ is
$y^2=-4x\cdots \left(ii\right)$


From equations $\left(i\right)$ and $\left(ii\right)$
$\frac{x^2}{9}-\frac{x}{2}=1\Rightarrow 2x^2-9x-18=0\Rightarrow (2x+3)(x-6)=0$
As the point of intersection is in second and third quadrant, so $x<0$,
$x=-\frac{3}{2}\Rightarrow y=\pm \sqrt{6}$
Therefore, the points of intersections are
$M\equiv (-\frac{3}{2},\sqrt{6}),N\equiv (-\frac{3}{2},-\sqrt{6})\Rightarrow d=MN=2\sqrt{6}\therefore d^2=24$