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A standard rectangular hyperbola has its vertices at $(5 , 7 )$ and $(-3 , -1 ) $ find its equation and asymptotes.

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Toolbox:
  • In a rectangular hyperbola the asymptotes are at right angles $a=b$ and $e=\sqrt 2$
  • http://clay6.com/mpaimg/1_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:
The midpoint of the line joining the vertices $(5,7),(-3,-1)$ is the centre of the hyperbola.
(i.e) $c\big(\large\frac{5-3}{2},\large\frac{7-1}{2}\big)$$=c(1,3)$
In the standard form (i.e asymptotes parallel to the coordinate axis),its equation is $(x-1)(y-3)=c^2$
Step 2:
Since the vertices lie on the hyperbola,$(5,7)$ is a point on the rectangular hyperbola.
$(5-1)(7-3)=c^2$
$c^2=16$
Therefore the equation of the rectangular hyperbola is $(x-1)(y-3)=16$
Step 3:
The combined equation of the asymptotes differs only in the constant.
Therefore $(x-1)(y-3)=k$
But the asymptotes pass through the centre $(1,3)$.
Therefore the combined equation of the asymptotes is $(x-1)(y-3)=0$ and the asymptotes are $x=1,y=3$
answered Jun 24, 2013 by sreemathi.v
edited Jun 24, 2013 by sreemathi.v
 

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