What is the vertices (y^2)/(25)-(x^2)/(64)=1 ?

The vertices (y^2)/(25)-(x^2)/(64)=1 is (0,5),(0,-5)

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vertices (y^2)/(25)-(x^2)/(64)=1

vertices

\frac{y ^{2}}{25} - \frac{x ^{2}}{64} = 1

(0, 5), (0, - 5)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{25} - \frac{x ^{2}}{64} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 5, b = 8

(0, 0 + 5), (0, 0 - 5)

Refine

(0, 5), (0, - 5)

9 y^{2} - 25 x^{2} = 225

2 x^{2} - y^{2} - 2 y = - 5

9 y^{2} + 36 y - 4 x^{2} - 24 x - 36 = 0

\frac{( x - 8 ) ^{2}}{4} - \frac{( y - 5 ) ^{2}}{9} = 1

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

What is the vertices (y^2)/(25)-(x^2)/(64)=1 ?
The vertices (y^2)/(25)-(x^2)/(64)=1 is (0,5),(0,-5)