First slide

Hyperbola in conic sections

Question

If $θ$ is an angle between the two asymptotes of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, then \cos (\frac{θ}{2})$ is equal to

Moderate

A

$\frac{b}{\sqrt{a^{2} + b^{2}}}$
B

$\frac{a b}{\sqrt{a^{2} + b^{2}}}$
C

$\frac{a}{\sqrt{a^{2} + b^{2}}}$
D

$\sqrt{\frac{a - b}{a + b}}$

Solution

Slopes of the asymptotes are $\pm \frac{b}{a}$
$\Rightarrow \tan θ = \frac{\frac{b}{a} - (- \frac{b}{a})}{1 + (\frac{b}{a}) (- \frac{b}{a})}$
$\Rightarrow \frac{2 \tan (\frac{θ}{2})}{1 - \tan^{2} (\frac{θ}{2})} = \frac{2 (b / a)}{1 - {(\frac{b}{a})}^{2}}$
$\begin{array}{l} \Rightarrow \tan (\frac{θ}{2}) = \frac{b}{a} as \tan (\frac{θ}{2}) > 0 \\ \Rightarrow \cos (\frac{θ}{2}) = \frac{1}{\sqrt{1 + \frac{b^{2}}{a^{2}}}} = \frac{a}{\sqrt{a^{2} + b^{2}}} \end{array}$

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