Question

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

Given ellipse is

$\frac{x^2}{9}+\frac{y^2}{8}=1-\left(1\right)$

Here $a^2=9,b^2=9$

$\therefore e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{8}{9}}=\frac{1}{3}$

So the cordinate of focci are $F_1(-1,0)and{F}_2(1,0).$

The equation of parabola having vertex at origin and focus at $F_2(1,0)is{y}^2=4x-\left(2\right)$

Solving equation (1) and (2) we get, cordinates of $M(\frac{3}{2},\sqrt{6})andN(\frac{3}{2},-\sqrt{6})$

Clearly altitude from F is a x-axis i.e.y=0.

Equation of Altitude from M is

$y-\sqrt{6}=\frac{5}{2\sqrt{6}}(x-\frac{3}{2})$

Solving equation with y=0 we get $(\frac{-9}{10},0)$ as orthocenter
ΔF1MN