Question

Let $F_1(x_1,0)$ and $F_2(x_2,0),$ 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. $(-\frac{9}{10},0)$
• B. $(\frac{2}{3},0)$
• C. $(\frac{9}{10},0)$
• D. $(\frac{2}{3},\sqrt{6})$

Answer 1

The correct option is A $(-\frac{9}{10},0)$


Equation of the ellipse:
$\frac{x^2}{9}+\frac{y^2}{8}=1\cdots \left(1\right)$
$a=3,b=2\sqrt{2}$
Hence, eccentricity is
$e^2=1-\frac{b^2}{a^2}=\frac{1}{9}\Rightarrow e=\frac{1}{3}$

So, foci of ellipse are $(\pm 1,0)$
Equation of the parabola having vertex $(0,0)$ and focus $(1,0)$ is
$y^2=4x\cdots \left(2\right)$

From equations $\left(1\right)$ and $\left(2\right),$ we get
$\frac{x^2}{9}+\frac{4x}{8}=1\Rightarrow 2x^2+9x-18=0\Rightarrow (2x-3)(x+6)=0\Rightarrow x=\frac{3}{2}[\because x=-6\to \text{not possible}]\therefore M\equiv (\frac{3}{2},\sqrt{6}),N\equiv (\frac{3}{2},-\sqrt{6})$

Equation of altitude from vertex $M(\frac{3}{2},\sqrt{6})$ is
$(y-\sqrt{6})=\frac{5}{2\sqrt{6}}(x-\frac{3}{2})$

$\because x-$axis is altitude drawn through vertex $F_1$
$\therefore$ orthocentre lies on the $x-$axis,
$\therefore 0-\sqrt{6}=\frac{5}{2\sqrt{6}}(x-\frac{3}{2})\Rightarrow x=-\frac{9}{10}$

Hence, the orthocentre of $△F_{1}MN$ is
$(-\frac{9}{10},0)$