Comment
Share
Q)

# The vertices of the hyperbola $9x^2-16y^2-36x+96y-252=0$ are

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

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 $x= X+2$
$y= Y+3$
$(x = \pm 4,y= 3)$
(i.e) $(x=6,y=3)$
$(x=-2,y=3)$