Question

Let $F_1(x_1,0)$ and $F_2(x_2,0),$ where $x_1<0$ 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 $\Delta F_{1}MN$ 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

$(-\frac{9}{10},0)$

here,$\frac{x^2}{9}+\frac{y^2}{8}=1$
has foci $(\pm ae,0)$
Where, $a^2{e}^2=a^2-b^2$
$\Rightarrow a^2{e}^2=9-8\Rightarrow ae=\pm 1$
i.e. $F_1,F_2$= $(\pm 1,0)$


Equation of parabola having vertex $O(0,0)$ and $F_2(1,0)$
$(as{x}^2>0)$
On solving $\frac{x^2}{9}+\frac{y^2}{8}=1$ and $y^2=4x$, we get
$x=\frac{3}{2}$ and $y=\pm \sqrt{6}$
slope of altitude through M on $N{F}_1$ is
$=\frac{5}{2\sqrt{6}}$[using slope of two perpendicular lines]
Using point slope form, the equation of the perpendicular
$\Rightarrow (y-\sqrt{6})=\frac{5}{2\sqrt{6}}(x-\frac{3}{2})$
And equation of altitude through $F_1$ is $y=0$
On solving these Eqs. , we get $(\frac{-9}{10},0)$