What is the vertices (y^2)/(25)-(x^2)/(49)=1 ?

The vertices (y^2)/(25)-(x^2)/(49)=1 is (0,5),(0,-5)

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vertices (y^2)/(25)-(x^2)/(49)=1

vertices

\frac{y ^{2}}{25} - \frac{x ^{2}}{49} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{25} - \frac{x ^{2}}{49} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 5, b = 7

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

Refine

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

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

- x^{2} + 4 y^{2} + 2 x + 24 y + 31 = 0

9 x^{2} - 4 y^{2} = 144

4 y^{2} - 9 x^{2} + 40 y - 540 x - 8036 = 0

11 x^{2} - 25 y^{2} + 22 x + 250 y - 889 = 0

What is the vertices (y^2)/(25)-(x^2)/(49)=1 ?
The vertices (y^2)/(25)-(x^2)/(49)=1 is (0,5),(0,-5)