Question

If a triangle formed by the latusrectum of a hyperbola with the farther vertex of the conic is an equilateral triangle, then the eccentricity is equal to

• A. $\sqrt{3}$
• B. $\sqrt{3}+1$
• C. $\frac{\sqrt{3}+1}{\sqrt{2}}$
• D. $\frac{\sqrt{3}+1}{\sqrt{3}}$

Answer 1

The correct option is D $\frac{\sqrt{3}+1}{\sqrt{3}}$

Length of a side of the equilateral triangle is$=$Length of latusrectum$=\frac{2b^2}{a}$
Length of height of the equilateral triangle$=$ Distance between one focus and vertex of conjugate hyperbola$=ae+a$
Height of the equilateral triangle$=\frac{\sqrt{3}}{2}\times side$
$\Rightarrow \frac{\sqrt{3}}{2}\left(\frac{2b^2}{a}\right)=a(1+e)$
$\Rightarrow \sqrt{3}{b}^2=a^2(1+e)$
$\Rightarrow \sqrt{3}{a}^2(e^2-1)=a^2(1+e)$
$\Rightarrow \sqrt{3}(e^2-1)=(1+e)$
$\Rightarrow (e-1)=\frac{1}{\sqrt{3}}$ (Since, $e\ne -1$)
$\Rightarrow e=1+\frac{1}{\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}}$
Hence, option D is correct.