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JEE Advanced · Mathematics · 14. Ellipse

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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 \(\frac{x^{2}}{9}+\frac{y^{2}}{8}=1 .\) 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.


Question:

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 \(M Q R\) to area of the quadrilateral \(M F_{1} N F_{2}\) is

  1. A 3 : 4
  2. B 4 : 5
  3. C 5 : 8
  4. D 2 : 3
Verified Solution

Answer & Solution

Correct Answer

(C) 5 : 8

Step-by-step Solution

Detailed explanation


Equation of chord of contact MN is x= 3 2
Let R(h,k)
Equation of chord of contact
xh 9 + yk 8 =1
Comparing we get R(6,0)
Normal to parabola at M :y- 6= -62·1 x-32
Solving it with y=0, we get Q 72, 0
Area of triangle MQR =12. 6-72. 6=564
Area of quadrilateral MF1NF2=2·121--16=26
Required ratio = 5 : 8
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