Question

If latus rectum of an ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1\{0<b<4\}$, subtends angle $2\theta$ at farthest vertex such that $\text{cosec}\theta =\sqrt{5}$ then which of the following options are correct:

• A. $e=\frac{1}{2}$
• B. no such ellipse exists
• C. $b=2\sqrt{3}$
• D. area of $\Delta$ formed by $LR$ and nearest vertex is $6sq$. units

Answer 1

Answer: (A), (C), (D)

The correct options are
A $e=\frac{1}{2}$
C $b=2\sqrt{3}$
D area of $\Delta$ formed by $LR$ and nearest vertex is $6sq$. units

$Given,a=4Distance,AS=\frac{b^2}{a}=\frac{b^2}{4}Distance,BS=a+aeBSbisectstheanglemadebythelatusrectumatthefarthestend.\therefore tan\theta =\frac{b^2}{4(a+ae)}But,\csc \theta =\sqrt{5}So,sin\theta =\frac{1}{\sqrt{5}}cos\theta =\frac{2}{\sqrt{5}}And,tan\theta =\frac{1}{2}Or,\frac{b^2}{4(a+ae)}=\frac{1}{2}Or,\frac{a^2(1-e^2)}{4a(1+e)}=\frac{1}{2}Or,1-e=\frac{1}{2}Or,e=\frac{1}{2}\therefore b=\sqrt{a^2(1-e^2)}=\sqrt{16\frac{3}{4}}=2\sqrt{3}AreaofthetriangleformedbyLRandnearestvertexis:A=\frac{1}{2}\ast \frac{2b^2}{a}\ast (a-ae)=\frac{12}{4}\ast 2=6sq.units$

Option [D][C][A]