What is the vertices (y^2)/(25)-(x^2)/(36)=1 ?

The vertices (y^2)/(25)-(x^2)/(36)=1 is (0,5),(0,-5)

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vertices (y^2)/(25)-(x^2)/(36)=1

vertices

\frac{y ^{2}}{25} - \frac{x ^{2}}{36} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{25} - \frac{x ^{2}}{36} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 5, b = 6

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

Refine

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

y^{2} - 4 x^{2} = 36

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{16} = 1

\frac{x ^{2}}{49} - y^{2} = 1

\frac{( y - 2 ) ^{2}}{4} - \frac{( x + 3 ) ^{2}}{9} = 1

\frac{x ^{2}}{81} - \frac{y ^{2}}{144} = 1

What is the vertices (y^2)/(25)-(x^2)/(36)=1 ?
The vertices (y^2)/(25)-(x^2)/(36)=1 is (0,5),(0,-5)