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Find the angle between the asymptotes of hyperbola $24x^{2}-8y^{2}=27$

1 Answer

Toolbox:
  • (i) The asymptotes of $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ are $\large\frac{x}{a}$$\pm\large\frac{y}{b}$$=0$.
  • (ii) They pass through the centre of the hyperbola.
  • (iii) Their slopes are $\large\frac{b}{a}$ and $-\large\frac{b}{a}$( i.e)the axes of the hyperbola bisect the angles between them.
  • (iv) The angle between the asymptotes $2\alpha=2\tan^{-1}\large\frac{b}{a}$$=2\sec^{-1}e$
Step 1:
$24x^2-8y^2=27$
The above equation is divided by $27$ we get
$\large\frac{x^2}{\Large\frac{27}{24}}-\large\frac{y^2}{\Large\frac{27}{8}}$$=1$
$\large\frac{x^2}{\Large\frac{9}{8}}-\large\frac{y^2}{\Large\frac{27}{8}}$$=1$
$a^2=\large\frac{9}{8}$,$b^2=\large\frac{27}{8}$
$\Rightarrow a=\large\frac{3}{2\sqrt 2},$$b=\large\frac{3\sqrt 3}{2\sqrt 2}$
Step 2:
Let $2\alpha$ be the angle between the asymptotes $2\alpha=2\tan^{-1}\large\frac{b}{a}$
$\Rightarrow 2\tan^{-1}\large\frac{\Large\frac{3\sqrt 3}{2\sqrt 2}}{\Large\frac{3}{2\sqrt 2}}$
$\Rightarrow 2\tan^{-1}\big(\large\frac{3\sqrt 3}{2\sqrt 2}\times \large\frac{2\sqrt 2}{3}\big)$
$\Rightarrow 2\tan^{-1}\sqrt 3$
$\Rightarrow 2\times \large\frac{\pi}{3}$
$\Rightarrow \large\frac{2\pi}{3}$ or $120^{\large\circ}$
answered Jun 24, 2013 by sreemathi.v
 

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