What is the vertices (y^2)/(16)-(x^2)/(25)=1 ?

The vertices (y^2)/(16)-(x^2)/(25)=1 is (0,4),(0,-4)

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vertices (y^2)/(16)-(x^2)/(25)=1

vertices

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

(0, 4), (0, - 4)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (0, 0), a = 4, b = 5

(0, 0 + 4), (0, 0 - 4)

Refine

(0, 4), (0, - 4)

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

12 x^{2} + 20 y^{2} - 12 x + 40 y - 37 = 0

What is the vertices (y^2)/(16)-(x^2)/(25)=1 ?
The vertices (y^2)/(16)-(x^2)/(25)=1 is (0,4),(0,-4)