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JEE Advance - Mathematics (2016 - Paper 2 Offline - No. 4)

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 $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$. 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 $$MQR$$ to area of the quadrilateral $$M{F_1}N{F_2}$$is

$$3:4$$
$$4:5$$
$$5:8$$
$$2:3$$

Explanation

Equation of tangent at M(3/2, $$\sqrt6$$) to $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$ is

$${3 \over 2}.{x \over 9} + \sqrt 6 .{y \over 8} = 1$$ ....... (i)

which intersect X-axis at (6, 0).

Also, equation of tangent at N(3/2, $$-$$$$\sqrt6$$) is

$${3 \over 2}.{x \over 9} - \sqrt 6 .{y \over 8} = 1$$ ....... (ii)

Eqs. (i) and (ii) intersect on X-axis at R(6, 0). ........ (iii)

Also, normal at $$M(3/2,\sqrt 6 )$$ is

$$y - \sqrt 6 = {{ - \sqrt 6 } \over 2}\left( {x - {3 \over 2}} \right)$$

On solving with y = 0, we get Q(7/2, 0) ....... (iv)

JEE Advanced 2016 Paper 2 Offline Mathematics - Ellipse Question 11 English Explanation

The area of MQR is

$$\left| {{1 \over 2}} \right|\left| {\matrix{ {3/2} & {\sqrt 6 } & 1 \cr 6 & 0 & 1 \cr {7/2} & 0 & 1 \cr } } \right| = \left| {{{\sqrt 6 } \over 2}\left( {6 - {7 \over 2}} \right)} \right| = {{5\sqrt 6 } \over 4}$$

The area of the quadrilateral MF1NF2 is

$$2(\Delta {m_1}{F_1}{F_2}) = 2\sqrt 6 $$

and the required ratio is

$${{5\sqrt 6 } \over {4\,.\,2\sqrt 6 }} = {5 \over 8}$$

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