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Find the angle between the asymptotes of hyperbola $9(x-2)^{2}-4(y+3)^{2}=36$

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  • (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:
$9(x-2)^2-4(y+3)^2=36$
The above equation is divided by $36$ we get
$\large\frac{(x-2)^2}{4}-\large\frac{(y+3)^2}{9}$$=1$
The hyperbola has its centre at $(2,-3)$ and $a^2=4,b^2=9$
$\Rightarrow a=2,b=3$
Step 2:
Let $2\alpha$ be the angle between the asymptotes.
$2\alpha=2\tan^{-1}\large\frac{b}{a}$
$\Rightarrow 2\tan^{-1}\large\frac{3}{2}$
The angle between the asymptotes is $2\tan^{-1}\large\frac{3}{2}$
answered Jun 24, 2013 by sreemathi.v
 

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