JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Ellipse

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.

If the tangents to the ellipse at

M

and

N

meet at

R

and the normal to the parabola at

M

meets the

x

-axis at

Q

, then the ratio of area of the triangle

M QR

to area of the quadrilateral

M F_{1} N F_{2}

Answer: (C)

Solution

The tangents to the ellipse at

M (\frac{3}{2}, 6)

and

N (\frac{3}{2}, - 6)

are

\frac{x}{6} \pm \frac{y 6}{8} = 1

. They meet at

R (6, 0)

. The normal to the parabola

y^{2} = 4 x

M

(y - 6) = - \frac{6}{2} (x - \frac{3}{2})

. Setting

y = 0

x = \frac{7}{2}

, so

Q (\frac{7}{2}, 0)

. Area of

△ M QR = \frac{1}{2} \times (6 - \frac{7}{2}) \times 6 = \frac{5 6}{4}

. The quadrilateral

M F_{1} N F_{2}

has area

\frac{1}{2} (F_{1} F_{2}) (y_{M} - y_{N}) = \frac{1}{2} \times 2 \times 26 = 26

. Ratio:

\frac{5 6 /4}{2 6} = \frac{5}{8}

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option