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Question

Let F1(x1,0) and F2(x2,0), for x1<0 and x2>0, be the foci of the ellipse x29+y28=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 F1MN is

A
(910,0)
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B
(23,0)
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C
(910,0)
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D
(23,6)
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Solution

The correct option is A (910,0)

Equation of the ellipse:
x29+y28=1 (1)
So, Foci of ellipse are (±1,0)
Equation of the parabola having vertex (0,0) and focus (1,0) is y2=4x (2)
From equation (1) and (2), we get
x29+4x8=12x2+9x18=0
x=32 [6 not possible]
M=(32,6) and N=(32,6)

Equation of altitude from vertex M(32,6) is
(y6)=526(x32)

According to the figure, orthocentre lies on xaxis,
06=526(x32)x=910
Hence, orthocentre of F1MN is (910,0)

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