How to get the foci / focus Hyperbola

Question

How to find the lower and upper focus? Hyperbola

I started with this $$ 9x^2 + 54x - y^2 + 10y + 81 = 0 $$

and broke it down to

$$ \frac{9(x+3)^2}{25} - \frac{(y-5)^2}{25} = -1 $$

center = (-3,5) Lower Vertex = (-3,0) Upper Vertex = (-3,10)

How to get the foci?

foci / focus = (h, k +- c)

b = 5 but what is a?

$$ c^2 = a^2 + b^2 $$

Thank you.

Answer 1

Hint: Write the equation as $$\dfrac {(x+3)^2}{\dfrac{25}{9}}-\dfrac{(y-5)^2}{25}=-1.$$

Answer 2

We have $a^2=\frac{25}{9}$ because we have $$\frac{(x+3)^2}{\dfrac{25}{9}}-\frac{(y-5)^2}{25}=-1.$$

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The foci of a hyperbola $$\frac{(x-m)^2}{a^2}-\frac{(y-n)^2}{b^2}=-1$$ are $$\left(m,n\pm\sqrt{a^2+b^2}\right).$$