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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

\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.

Section: Mathematics

Question : 10 of 25

Marks: +1, -0

MF_1NF_2

Solution:

Normal to parabola at M:

y-\sqrt{6}=-\frac{\sqrt{6}}{2.1}\left(x-\frac{3}{2}\right)

Solving it with y = 0, we get

Q=\left(\frac{7}{2},0\right)

Tangent to ellipse at M:

\frac{x \cdot \frac{3}{2}}{9}+\frac{y\sqrt{6}}{8}=1

solving it with y=0,we get R≡ (6,0)

∴ Area of triangle MQR

=\frac{1}{2} \cdot \left(6-\frac{7}{2}\right) \cdot \sqrt{6} = \frac{5\sqrt{6}}{4}

Area of quadrilateral

MF_1NF_2=2 \cdot \frac{1}{2} \cdot (1-(-1)) \cdot \sqrt{6}=2\sqrt{6}

Required ratio = 5 : 8

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