Paragraph: Let $F_{1}\left(x_{1}, 0\right)$ and $F_{2}\left(x_{2}, 0\right)$, for $x_{1}<0$ and $x_{2}>0$, be the foci of the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{8}=1 .$ Suppose a parabola having vertex at the origin and focus at $F_{2}$ intersects the ellipse at point $M$ in the first quadrant and at point $N$ in the fourth quadrant.
Question: The orthocentre of the triangle $F_{1} M N$ is

• Correct answer
  A

  $\left(-\frac{9}{10}, 0\right)$
• B

  $\left(\frac{2}{3}, 0\right)$
• C

  $\left(\frac{9}{10}, 0\right)$
• D

  $\left(\frac{2}{3}, \sqrt{6}\right)$

Solution

$\frac{x^2}{9}+\frac{y^2}{8}=1$
$F_1\left(-1,0\right),F_2\left(1,0\right)$
$\Rightarrow$ Parabola is $y^2=4x$
The intersection of ellipse & parabola is $M \& N$
$\frac{x^2}{9}+\frac{4x}{8}=1\Rightarrow M\left(\frac{3}{2},\sqrt{6}\right)\&N\left(\frac{3}{2},-\sqrt{6}\right)$

Equation of altitude through $M :y- \sqrt{6}=\frac{5}{2\sqrt{6}} \left(x-\frac{3}{2}\right)$
Equation of altitude through $F_1:y=0$
Solving, we get orthocentre $\left(-\frac{9}{10}, 0\right)$

Hence, the correct answer is option: 'A'.