Question

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

• A. $(6,3)$ and $(-6,3)$
• B. $(6,3)$ and $(-2,3)$
• C. $(-6,3)$and$(-6,-3)$
• D. None of these

Answer 1

The correct option is B $(6,3)$ and $(-2,3)$

We have, $9(x^2-4x+4)-16(y^2-6y+9)=144$
$\Rightarrow$ $\frac{(x-2{)}^2}{4^2}-\frac{(y-3{)}^2}{3^2}=1$
Shifting the origin at (2, 3), we have $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ Where, x = X + 2, y = Y + 3. Then coordinates of the vertices are $x=\pm 4,Y=3$ i.e. $x=6,y=3$ and $x=-2,y=3$