JEE Advanced 2016 · Paper 2 · Mathematics Q53

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MathematicsSingle correct+3Question 53 of 54

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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 $\dfrac{x^2}9+\dfrac{y^2}8=1$ . 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 $F_1MN$ is

\left(-\dfrac9{10},0\right)

\left(\dfrac23,0\right)

\left(\dfrac9{10},0\right)

\left(\dfrac23,\sqrt6\right)

Answer and worked solution

Official answer: A

Foci. $c^2=9-8=1$ , so $F_1=(-1,0)$ and $F_2=(1,0)$ . The parabola is $y^2=4x$ .

Intersection. $\dfrac{x^2}9+\dfrac x2=1$ gives $x=\dfrac32$ , so $M=\left(\frac32,\sqrt6\right)$ and $N=\left(\frac32,-\sqrt6\right)$ .

Orthocentre. By symmetry it is $(h,0)$ . The altitude from M is perpendicular to $F_1N$ , which has slope $-\dfrac{2\sqrt6}5$ :
$0-\sqrt6=\frac5{2\sqrt6}\left(h-\frac32\right)\implies h=\frac32-\frac{12}5=-\frac9{10}.$

(A)

Topics:Ellipse Parabola

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JEE Advanced 2016 · Paper 2, Mathematics question 53