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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

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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.x9−√6y8=1

Both the tangents intersects x-Axis at R (6,0)

Also normal to the parabola at M (32,√6) is y−√6=−√62(x−32)

On solving with y=0, we get (72,0)

∴area of ΔMQR=12(6−72)√6=5√64sq units

and area of quadrilateral MF1NF2=2×(12(1−(−1))√6)=2√6 sq units

∴area of ΔMQRArea of quadrilateral M1NF2=58

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