Verified Solution • 2025–26 Session moderate

QJEEAdv16Ma96b4: JEE Advanced 2016 Mathematics - Class 11 JEEAdv16Ma96b4 Solution: Let and for , be the foci of the ellipse . Suppose a parabola…

Let F1(x1,0){F_1}\left( {{x_1},0} \right) and F2(x2,0){F_2}\left( {{x_2},0} \right) for {{x_1} < 0} and {{x_2} > 0} , be the foci of the ellipse x29+y28=1{{{x^2}} \over 9} + {{{y^2}} \over 8} = 1 . Suppose a parabola having vertex at the origin and focus at F2{F_2} intersects the ellipse at point MM in the first quadrant and at point NN in the fourth quadrant.

If the tangents to the ellipse at MM and NN meet at RR and the normal to the parabola at MM meets the xx -axis at QQ , then the ratio of area of the triangle MQRMQR to area of the quadrilateral MF1NF2M{F_1}N{F_2} is

Multiple Choice Options

  • A 3:43:4
  • B 4:54:5
  • C 5:85:8
  • D 2:32:3

Step-by-Step Verified Solution

Equation of tangent at M(3/2, 6\sqrt6 ) to x29+y28=1{{{x^2}} \over 9} + {{{y^2}} \over 8} = 1 is

32.x9+6.y8=1{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, −- 6\sqrt6 ) is

32.x9−6.y8=1{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(3/2,6)M(3/2,\sqrt 6 ) is

y−6=−62(x−32)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

|12||3/2616017/201|=|62(6−72)|=564

The area of the quadrilateral MF1NF2 is

2(Δm1F1F2)=262(\Delta {m_1}{F_1}{F_2}) = 2\sqrt 6

and the required ratio is

564 . 26=58{{5\sqrt 6 } \over {4\,.\,2\sqrt 6 }} = {5 \over 8}

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring