Statement-I : If a latusrectum of an ellipse subtends angle 60° at the farthest vertex then eccentricity is 1-13Statement-II : If a latusrectum subtends 60° at the centre of the ellipse then ecentricity is 1+2323which of the above statements are true

(I)13=b2a(a+ae)=a2(1-e2)a2(e+1) 13=1-e ⇒ e=1-13 (II)13=b2a2e=a2(1-e2)a2e e=3-3 e2 3 e2+e-3=0 e=-1±1+1223 e=-1+1323

Statement-I : If a latusrectum of an ellipse subtends angle 60° at the farthest vertex then eccentricity is 1-13Statement-II : If a latusrectum subtends 60° at the centre of the ellipse then ecentricity is 1+2323which of the above statements are true

(I)13=b2a(a+ae)=a2(1-e2)a2(e+1) 13=1-e ⇒ e=1-13 (II)13=b2a2e=a2(1-e2)a2e e=3-3 e2 3 e2+e-3=0 e=-1±1+1223 e=-1+1323

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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

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only I

only II

both I & II

neither I nor II

answer is A.

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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)

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