What is the vertices (y^2)/(225)-(x^2)/(64)=1 ?

The vertices (y^2)/(225)-(x^2)/(64)=1 is (0,15),(0,-15)

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vertices (y^2)/(225)-(x^2)/(64)=1

vertices

\frac{y ^{2}}{225} - \frac{x ^{2}}{64} = 1

(0, 15), (0, - 15)

Solution steps

(h, k + a), (h, k - a)

Calculate Hyperbola properties

\frac{y ^{2}}{225} - \frac{x ^{2}}{64} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 15, b = 8

(0, 0 + 15), (0, 0 - 15)

Refine

(0, 15), (0, - 15)

\frac{y ^{2}}{1} - \frac{x ^{2}}{36} = 1

8 y^{2} - 4 x^{2} = 4

9 x^{2} - y^{2} + 18 x + 8 y = 43

\frac{( x - 7 ) ^{2}}{36} - \frac{( y - 3 ) ^{2}}{64} = 1

16 x^{2} - 3 y^{2} - 32 x + 12 y - 44 = 0

What is the vertices (y^2)/(225)-(x^2)/(64)=1 ?
The vertices (y^2)/(225)-(x^2)/(64)=1 is (0,15),(0,-15)