Question

What are the vertices of the hyperbola given by the equation y2 – 9x2 + 22y – 72x – 32 = 0?

(-3, -11) and (-5, -11)

(-1, -11) and (-7, -11)

(-4, -8) and (-4, -14)

(-4, -10) and (-4, -12)

Answer 1

had u tried to take it to standar form
by completing the square?

Answer 2

$$
y^2-9x^2+22y-72x-32=0\\
(y^2+22y)-(9x^2-72x)-32=0\\
(y^2+22y)-9(x^2-8x)-32=0\\
\text{now completing the perfect square trinomials}\\
(y^2+22y+\boxed{121})-9(x^2-8x+\boxed{16})-32=0
$$
can you take it from there?

Answer 3

$$
y^2-9x^2+22y-72x-32=0\\
(y^2+22y)-(9x^2-72x)-32=0\\
(y^2+22y)-9(x^2-8x)-32=0\\
\text{now completing the perfect square trinomials}\\
(y^2+22y+\boxed{121})-9(x^2-8x+\boxed{16})=32+\boxed{121}-9\times\boxed{16}
$$
rather