What is the vertices (y^2}{25}-\frac{x^2)/4 =1 ?

The vertices (y^2}{25}-\frac{x^2)/4 =1 is (0,5),(0,-5)

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vertices (y^2}{25}-\frac{x^2)/4 =1

vertices

\frac{y ^{2}}{25} - \frac{x ^{2}}{4} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{25} - \frac{x ^{2}}{4} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 5, b = 2

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

Refine

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

\frac{y ^{2}}{16} - \frac{x ^{2}}{25} = 1

x^{2} = - 16 y

x^{2} - 6 x - 4 y + 5 = 0

\frac{x ^{2}}{81} + \frac{y ^{2}}{49} = 1

x^{2} + y^{2} = 20

What is the vertices (y^2}{25}-\frac{x^2)/4 =1 ?
The vertices (y^2}{25}-\frac{x^2)/4 =1 is (0,5),(0,-5)