JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 20)

Let S and S' be the foci of an ellipse and B be any one of the extremities of its minor axis. If $$\Delta $$S'BS is a right angled triangle with right angle at B and area ($$\Delta $$S'BS) = 8 sq. units, then the length of a latus rectum of the ellipse is :

4$$\sqrt 2 $$

2$$\sqrt 2 $$

Explanation

b² = a²e² . . . . . . (i)

$${1 \over 2}$$ S'B.SB = 8

S'B.SB = 16

a²e² + b² = 16 . . . . .(ii)

b² = a² (1 $$-$$ e²) . . . . .(iii)

using (i), (ii), (iii) a = 4

      b = $$2\sqrt 2 $$

      e = $${1 \over {\sqrt 2 }}$$

$$ \therefore $$  $$\ell $$ (L.R) $$=$$ $${{2{b^2}} \over a} = 4$$

JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 20)

Explanation

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