What is the vertices y^2-25x^2+8y-9=0 ?

The vertices y^2-25x^2+8y-9=0 is (0,1),(0,-9)

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vertices y^2-25x^2+8y-9=0

vertices

y^{2} - 25 x^{2} + 8 y - 9 = 0

(0, 1), (0, - 9)

Solution steps

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

Calculate Hyperbola properties

y^{2} - 25 x^{2} + 8 y - 9 = 0 :

Up-down Hyperbola with

(h, k) = (0, - 4), a = 5, b = 1

(0, - 4 + 5), (0, - 4 - 5)

Refine

(0, 1), (0, - 9)

x^{2} - 16 y^{2} = 144

25 x^{2} - 16 y^{2} + 50 x - 32 y - 391 = 0

\frac{x ^{2}}{49} - \frac{( y + 6 ) ^{2}}{4} = 1

5 y^{2} - 16 x^{2} + 400 = 0

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

What is the vertices y^2-25x^2+8y-9=0 ?
The vertices y^2-25x^2+8y-9=0 is (0,1),(0,-9)