Question

For a hyperbola, angle between the asymptotes is ${60}^{\circ },$ and the length of transverse axis is $24$ units, then

• A. $e=\frac{2}{\sqrt{3}}$
• B. $e=\frac{4}{\sqrt{3}}$
• C. Equation of hyperbola:
  $\frac{x^2}{144}-\frac{y^2}{36}=1$
• D. Equation of hyperbola:
  $\frac{x^2}{144}-\frac{y^2}{48}=1$

Answer 1

Answer: (A), (D)

Equation of hyperbola:

$\frac{x^2}{144}-\frac{y^2}{48}=1$
Given: angle between the asymptote is ${60}^{\circ }$
and length of transverse axis that is $2a=24$

$e=\sec (\frac{\theta }{2})$
$\Rightarrow e=\sec (\frac{{60}^{\circ }}{2})$
$\Rightarrow e=\frac{2}{\sqrt{3}}...\left(i\right)$

also,
$e=\sqrt{1+\frac{b^2}{a^2}}$
$\Rightarrow \frac{2}{\sqrt{3}}=\sqrt{1+\frac{b^2}{(12{)}^2}}\left[\text{from}\right(i\left)\right]$
$\Rightarrow \frac{4}{3}=1+\frac{b^2}{(12{)}^2}\left[\text{from}\right(i\left)\right]$
$\Rightarrow b^2=48$

Equation of Hyperbola:
$\Rightarrow \frac{x^2}{144}-\frac{y^2}{48}=1$