Exam

JEE Advanced

Year

2016

Paper

Paper 2

Subject

Mathematics

Chapter

Straight Lines

JEE Advanced 2016 · Paper 2 · Mathematics · Straight Lines

Paragraph: Let F_ 1 (x_ 1 , 0) and F_ 2 (x_ 2 , 0) , for x_ 1 0 ,...

Paragraph: 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$in the fourth quadrant. <br> <strong> Question: </strong> The orthocentre of the triangle$F_{1} M N is

Step-by-step solution

$\dfrac{{x}^{2}}{9} + \dfrac{{y}^{2}}{8} = 1$
${F}_{1} ( - 1, 0 ), {F}_{2} ( 1, 0 )$
$⇒$ Parabola is ${y}^{2} = 4x$
The intersection of ellipse & parabola is $M \text{ and } N$
$\dfrac{{x}^{2}}{9} + \dfrac{4x}{8} = 1 ⇒ M( \dfrac{3}{2} , \sqrt{6} ) \text{ and } N( \dfrac{3}{2} , - \sqrt{6} )$

Equation of altitude through $M: y - \sqrt{6} = \dfrac{5}{2\sqrt{6}}\left(x - \dfrac{3}{2}\right)$
Equation of altitude through ${F}_{1}: y = 0$
Solving, we get orthocentre \left(- \dfrac{9}{10}, 0\right)