Question

PARAGRAPH 2
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.

The orthocentre of the triangle $F_{1}MN$ is

• A. $\left(-\frac{9}{10},0\right)$
• B. $\left(\frac{2}{3},0\right)$
• C. $\left(\frac{9}{10},0\right)$
• D. $\left(\frac{2}{3},\sqrt{6}\right)$

Answer 1

Answer: (A)

$\frac{x^2}{9}+\frac{y^2}{8}=1$
foci $=(\pm ae,0)$
$=\left(\pm 3\sqrt{1-\frac{8}{9}},0\right)$
$=(\pm 1,0)$
Equation of parabola
$y^2=4ax$
$y^2=4x$
intersection points of both curves are
$M\left(\frac{3}{2},\sqrt{6}\right),N\left(\frac{3}{2},-\sqrt{6}\right)$
Let the orthocentre be $(h,0)$.
$\therefore \left(\frac{\sqrt{6}-0}{h-\frac{3}{2}}\right)\left(\frac{0-(-\sqrt{6})}{\frac{3}{2}-(-1)}\right)=-1$
$\Rightarrow \frac{6}{\left(h-\frac{3}{2}\right)\left(\frac{5}{2}\right)}=-1$
$\Rightarrow 24=-10h+15$
$\Rightarrow h=-\frac{9}{10}$
$\therefore$ orthocentre $\left(-\frac{9}{10},0\right)$