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

Engineering

Mathematics

Standard Hyperbola

Question

$\frac{\sqrt{3} + \sqrt{5}}{2}$

$\frac{\sqrt{3} + 1}{2}$

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

$\sqrt{2}$

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$∵$    Slope of OL = 1
⇒ $ae = \frac{b^{2}}{a}$
⇒    a²e = a² (e² – 1)
⇒    e² – e – 1 = 0
⇒    $e = \frac{1 \pm \sqrt{5}}{2}$
∴   eccentricity = $\frac{\sqrt{5} + 1}{2}$