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Question

Let F1(x1,0) and F2(x2,0), where, x1<0 and x2>0 be the foci of the ellipse x29+y28=1 suppose a parabola having vertex at the origin and focus at F2 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 ΔMQR to area of the quadrilateral MF1NF2 is


A

3:4

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B

4:5

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C

5:8

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D

2:3

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Solution

The correct option is C

5:8


Equation of tangent at M(32,6) to x29+y28=1 is
32.x9+6y8=1
Which intersect X-axis at(6,0).
Also equation of tangent at N (32,6) is
32.x96y8=1
Both the tangents intersects x-Axis at R (6,0)
Also normal to the parabola at M (32,6) is y6=62(x32)
On solving with y=0, we get (72,0)

area of ΔMQR=12(672)6=564sq units
and area of quadrilateral MF1NF2=2×(12(1(1))6)=26 sq units
area of ΔMQRArea of quadrilateral M1NF2=58


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