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

Question

Aicharya · Accepted Answer

Given hyperbola is
$\begin{aligned}
& 9 x^2-16 y^2-36 x+96 y-252=0 \\
& \Rightarrow 9\left(x^2-4 x\right)-16\left(y^2-6 y\right)=252 \\
& \Rightarrow 9\left(x^2-4 x+4\right)-16\left(y^2-6 y+9\right) \\
& =252+36-144 \\
& \Rightarrow 9(x-2)^2-16(y-3)^2=144
\end{aligned}$

Put $x=X+2$ and $y=Y+3$
$\therefore$ Equation (i) becomes $\frac{X^2}{16}-\frac{Y^2}{9}=1$
Now, vertices are $X= \pm a$ wehre $a=4$ and $Y=0$
Hence, vertices are $(6,3),(-2,3)$.

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

Solution