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

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

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vertices y^2-x^2= 1/25

vertices

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

(0, \frac{1}{5}), (0, - \frac{1}{5})

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (0, 0), a = \frac{1}{5}, b = \frac{1}{5}

(0, 0 + \frac{1}{5}), (0, 0 - \frac{1}{5})

Refine

(0, \frac{1}{5}), (0, - \frac{1}{5})

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

4 x^{2} - 5 y^{2} + 20 = 0

9 (x - 3)^{2} - 4 (y - 1)^{2} = - 36

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

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

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