1. Tardigrade
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  4. PARAGRAPH 2 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 ( x 2/9)+( y 2/8)=1 . Suppose a parabola having vertex at the origin and focus at F2 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

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Q. 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

JEE AdvancedJEE Advanced 2016

Solution:

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$\frac{x^{2}}{9} +\frac{y^{2}}{8}=1 $
foci $=(\pm a e, 0) $
$=\left(\pm 3 \sqrt{1-\frac{8}{9}}, 0\right)$
$=(\pm 1,0)$
Equation of parabola
$y^{2}=4 a x $
$y^{2}=4 x$
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 $( h , 0)$.
$\therefore \left(\frac{\sqrt{6}-0}{h-\frac{3}{2}}\right)\left(\frac{0-(-\sqrt{6})}{\frac{3}{2}-(-1)}\right)=-1$
$\Rightarrow \frac{6}{\left(h-\frac{3}{2}\right)\left(\frac{5}{2}\right)}=-1$
$\Rightarrow 24=-10 h+15$
$\Rightarrow h =-\frac{9}{10}$
$\therefore $ orthocentre $\left(-\frac{9}{10}, 0\right)$