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Find the angle between the asymptotes of hyperbola $4x^{2}-5y^{2}-16x+10y+31=0$

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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:
$4x^2-5y^2-16x+10y+31=0$
Completing the squares,
$4(x^2-4x+4)-5(y^2-2y+1)=-31+16-5$
$4(x-2)^2-5(y-1)^2=-20$
The above equation is divided by $-20$ we get,
$\large\frac{-(x-2)^2}{5}+\large\frac{(y-1)^2}{4}$$=1$
Step 2:
$\large\frac{(y-1)^2}{4}+\large\frac{(x-2)^2}{5}$$=1$
The centre is at $(2,1)$ and $a^2=4,b^2=5$
$\Rightarrow a=2,b=\sqrt 5$
Let $2\alpha$ be the angle between the asymptotes.
$2\alpha=2\tan^{-1}\large\frac{b}{a}$
$\Rightarrow 2\tan^{-1}\large\frac{\sqrt 5}{2}$
The angle between the asymptotes is $2\tan^{-1}\large\frac{\sqrt 5}{2}$
answered Jun 24, 2013 by sreemathi.v
 

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