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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The vertices of the hyperbola $9x^2-16y^2-36x+96y-252=0$ are

$\begin{array}{1 1}(A)\;(6,3)\;and\; (-6,3) \\(B)\;(6,3)\;and\;(-2,3) \\(C)\;(-6,3)\;and\;(-6,-3) \\(D)\;\text{none of these} \end{array}$

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1 Answer

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We have $9(x^2-4x+4)-16(y^2-6y+9)=144$
=> $\large\frac{(x-2)^2}{4^2} -\frac{(y-3)^2}{3^2}$$=1$
Shifting the origin at $(2,3)$ we have
$\large\frac{x^2}{4^2}-\frac{y^2}{3^2}$$=1$
Where $x= X+2$
$y= Y+3$
Then coordinate of the vertices are
$(x = \pm 4,y= 3)$
(i.e) $(x=6,y=3)$
$(x=-2,y=3)$
Hence B is the correct answer.
answered Apr 11, 2014 by meena.p
 

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