Show that angle between the two asymptotes of a hyperbola x2a2 - y2b2 - 1 is 2 tan-2ba or 2 -1 e-

Asymptotes of given hyperbola are : y=±bax slope of 1st asymptote b=is m_1=ba similarly m_2= ba Angle between asymptotes #160; = tanθ=m_1 m ...

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Standard XII

Mathematics

Domain and Range of Basic Inverse Trigonometric Functions

Show that ang...

Question

Show that angle between the two asymptotes of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $2 {tan}^{- 2} \frac{b}{a}$ (or) $2 {sec}^{- 1} (e)$ .

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Solution

Asymptotes of given hyperbola are : $y = \pm \frac{b}{a} x$

slope of 1st asymptote b=is $m_{1} = \frac{b}{a}$
similarly $m_{2} = - \frac{b}{a}$

Angle between asymptotes = $tan θ = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}$

$\Rightarrow tan θ = \frac{\frac{2 b}{a}}{1 - \frac{b^{2}}{a^{2}}}$

$θ = 2 {tan}^{- 1} (\frac{b}{a})$

or $θ = 2 {sec}^{- 1} (e)$

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Show that angle between the two asymptotes of a hyperbola x2a2−y2b2=1 is 2tan−2ba (or) 2sec−1(e).

Asymptotes of given hyperbola are : y=±baxslope of 1st asymptote b=is m1=basimilarly m2=−baAngle between asymptotes =tanθ=m1−m21+m1m2⇒tanθ=2ba1−b2a2θ=2tan−1(ba)or θ=2sec−1(e)

Show that angle between the two asymptotes of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $2 {tan}^{- 2} \frac{b}{a}$ (or) $2 {sec}^{- 1} (e)$ .