First slide

Hyperbola in conic sections

Question

If the latus rectum of a hyperbola subtend an angle of 60° at the other focus, then eccentricity of the hyperbola is

Moderate

A

2
B

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

$2 \sqrt{3}$
D

$\sqrt{3}$

Solution

Let the equation of the hyperbola be
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$
$F_{1} (- a e, 0), F_{2} (a e, 0)$ be the foci, $e$ being the eccentricity.
Let $L F_{2} L^{'},$ the latus rectum subtend an angle of 60° at $F_{1},$
then slope of $L F_{1}$ is $\tan 30^{0} .$

So $\tan 30^{\circ} = \frac{b^{2} / a}{2 a e}$
$\begin{array}{l} \Rightarrow \frac{1}{\sqrt{3}} = \frac{a^{2} (e^{2} - 1)}{2 a^{2} e} = \frac{e^{2} - 1}{2 e} \\ \Rightarrow e^{2} - 1 = \frac{2}{\sqrt{3}} e \Rightarrow {(e - \frac{1}{\sqrt{3}})}^{2} = 1 + \frac{1}{3} \\ \Rightarrow e - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \Rightarrow e = \sqrt{3} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App