Statement-I : If a latusrectum of an ellipse subtends angle 60° at the farthest vertex then eccentricity is $1 - \frac{1}{\sqrt{3}}$
Statement-II : If a latusrectum subtends 60° at the centre of the ellipse then ecentricity is $\frac{1 + 2 \sqrt{3}}{2 \sqrt{3}}$
which of the above statements are true

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

neither I nor II

both I & II

only II

only I

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

(I)

$\frac{1}{\sqrt{3}} = \frac{b^{2}}{a (a + ae)} = \frac{a^{2} (1 - e^{2})}{a^{2} (e + 1)} \frac{1}{\sqrt{3}} = 1 - e \Rightarrow e = 1 - \frac{1}{\sqrt{3}}$

(II)

$\frac{1}{\sqrt{3}} = \frac{b^{2}}{a^{2} e} = \frac{a^{2} (1 - e^{2})}{a^{2} e} e = \sqrt{3} - \sqrt{3} e^{2} \sqrt{3} e^{2} + e - \sqrt{3} = 0 e = \frac{- 1 \pm \sqrt{1 + 12}}{2 \sqrt{3}} e = \frac{- 1 + \sqrt{13}}{2 \sqrt{3}}$