Question

What is the center of the hyperbola whose equation is (y+3)^2/81-(x-6)^2/89=1?

Answer 1

We have been given an equation of hyperbola $\frac{(y+3)^2}{81}-\frac{(x-6)^2}{89}=1$. We are asked to find the center of hyperbola.  

We know that standard equation of a vertical hyperbola is in form $\frac{(y-k)^2}{b^2}-\frac{(x-h)^2}{a^2}=1$, where point (h,k) represents center of hyperbola.

Upon comparing our given equation with standard vertical hyperbola, we can see that the value of h is 6.

To find the value of k, we need to rewrite our equation as:

$\frac{(y-(-3))^2}{81}-\frac{(x-6)^2}{89}=1$

Now we can see that value of k is $-3$. Therefore, the vertex of given hyperbola will be at point $(6,-3)$ and option D is the correct choice.