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

Correct answer: C. 4 Given that, ∠SB S ^ ' = (π)/(2) ⇒ ∠OBS = (π)/(4) ⇒ b = ae . . . (i) Also, area (∆ SB S ^ ' ) = 8 ⇒ (1)/(2) . 2ae . b = 8 ⇒ aeb = 8 . . . (ii) From equations (i) and (ii) , b ^ 2 = 8 and a ^ 2 = b ^ 2 + a ^ 2 e ^ 2 = 8 + 8 = 16 ⇒ b = 2 2 and a = 4 Thus, the length of the latus rectum is 2 b ^ 2 a = (16)/(4) = 4

JEE Main 201912 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Ellipse

JEE Main 2019 · 12 January · Evening Shift · Mathematics · Ellipse

Let S and S ^ ' be the foci of an ellipse and B be any one of the...

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 \left(\Delta{S}^{'}BS\right) = 8sq . units

, then the length of a latus rectum of the ellipse is : Let S and S ^ ' be the foci of an ellipse and B be any one of the extremities of its minor axis. If Δ S ^ ' BS is a

Options

A $2\sqrt{2}$
B $2$
C $4$
D $4\sqrt{2}$

Correct answer

C. $4$

Step-by-step solution

Given that,

Let S and S ^ ' be the foci of an ellipse and B be any one of the extremities of its minor axis. If Δ S ^ ' BS is a

$∠SB{S}^{'} = \dfrac{\pi}{2}$

$⇒ ∠OBS = \dfrac{\pi}{4}$

$⇒ b = ae . . . \left(i\right)$

Also, area \left(∆ SB{S}^{'}\right) = 8 $$ ⇒ \dfrac{1}{2} . 2ae . b = 8 $$ ⇒ aeb = 8 . . . \left(ii\right) $From equations \left(i\right) \text{ and } \left(ii\right)$ ,

${b}^{2} = 8$ and ${a}^{2} = {b}^{2} + {a}^{2}{e}^{2} = 8 + 8 = 16$

$⇒ b = 2\sqrt{2}$ and $a = 4$

Thus, the length of the latus rectum is $\dfrac{2{b}^{2}}{a} = \dfrac{16}{4} = 4$

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