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Find the equation of the asymptotes of the following rectangular hyperbola $2xy+3x+4y+1=0$

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Toolbox:
  • http://clay6.com/mpaimg/3_toolbox%2021.jpg
  • For the standard rectangular hyperbola,the coordinate axes are taken to the asymptotes and the equation of the hyperbola is $xy=c^2$ .
  • If the centre is at $(h,k)$ with asymptotes parallel to the coordinate axes,the equation is $(x-h)(y-k)=c^2$
Step 1:
$2xy+3x+4y+1=0$
The above equation is divided by $2$ we get,
$xy+\large\frac{3}{2}$$x+2y+\large\frac{1}{2}=$$0$
Step 2:
Now we take the LHS of the above equation
LHS=$x(y+\large\frac{3}{2})$$+2(y+\large\frac{3}{2})$$-3+\large\frac{1}{2}$
$\qquad=(x+2)(y+\large\frac{3}{2})-\large\frac{5}{2}$
$\Rightarrow (x+2)(y+\large\frac{3}{2})-\large\frac{5}{2}$$=0$
$(x+2)(y+\large\frac{3}{2})=\large\frac{5}{2}$
Step 3:
Since the equation is in standard form,the equations of the asymptotes are $x+2=0,y+\large\frac{3}{2}$$=0$
answered Jun 25, 2013 by sreemathi.v
 

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