Question

If the latus rectum subtends a right angle at the centre of a hyperbola, then its eccentricity is ?

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

Answer 1

The correct option is A $\sqrt{2}$

Consider the ellipse$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$

Point F is the focus on the positive X axis and AB is the latus rectum.

The coordinates of F are $(ae,0)$ and the coordinates of the ends of the latus rectum,AF are BF are $(ae,\frac{b^2}{a})and(ae,\frac{{-b}^2}{a})$ respectively.

It is given that the latus rectum subtends an angle of ${90}^o$ at the centre.

$\angle AOF={45}^o$

since $△AOB\cong \angle BOF$

$\angle OAF={45}^o$ since$\angle OFA={90}^o$

$OF=AF$

But $OF=ae$

$AF=\frac{b^2}{a}=ae.$

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

$e=\sqrt{2}$