Let $H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, where $ab0$, be a hyperbola in the $x y$-plane

Question

Let $H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, where $a>b>0$, be a hyperbola in the $x y$-plane whose conjugate axis $L M$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle LMN be $4 \sqrt{3}$. \The correct option is:

Professor PC Thomas Classes · Answer

(b) : Given, the area of\ $	riangle L M N=4 \sqrt{3}$ sq. units\ $\Rightarrow \frac{1}{2}(2 b)(\sqrt{3} b)=4 \sqrt{3} \Rightarrow b^{2}=4 \quad 	herefore b=2$\ As $\frac{a}{b}=	an 60^{\circ} \Rightarrow a=2 \sqrt{3}$\ Also, $b^{2}=a^{2}\left(e^{2}-1ight)$\ $\Rightarrow \quad 4=12\left(e^{2}-1ight) \Rightarrow e^{2}=1+\frac{1}{3}=\frac{4}{3}$\ $\Rightarrow e=\frac{2}{\sqrt{3}}$\ Now, $2 a e=2 	imes 2 \sqrt{3} 	imes \frac{2}{\sqrt{3}}=8$\ $	herefore \quad$ Length of latus rectum\ $=\frac{2 b^{2}}{a}=\frac{2 \cdot 4}{2 \sqrt{3}}=\frac{4}{\sqrt{3}}$\