Question

The latusrectum of a hyperbola subtends a right angle at its centre,then its e =

• A. $\frac{\sqrt{5}-1}{2}$
• B. $\frac{\sqrt{5}+1}{2}$
• C. $\frac{\sqrt{5}+2}{2}$
• D. $\frac{\sqrt{5}+1}{3}$

Answer 1

The correct option is B

$\frac{\sqrt{5}+1}{2}$

Tan ${45}^{\circ }$ = 1 = $\frac{\frac{b^2}{a}}{ae}$

$\frac{b^2}{a^{2}e}=1$

$\Rightarrow$ $b^2=a^{2}e$

$\Rightarrow$ $a^2$ ($e^2-1)$ = $a^2$ e

$\Rightarrow$ $e^2-e-1=0$ $\Rightarrow$ e = $\frac{1\pm \sqrt{1+4}}{2}=\frac{\sqrt{5}+1}{2}$