What is the vertices 4y^2-5x^2=20 ?

The vertices 4y^2-5x^2=20 is (0,sqrt(5)),(0,-sqrt(5))

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vertices 4y^2-5x^2=20

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

4 y^{2} - 5 x^{2} = 20 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 5, b = 2

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

Refine

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

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

(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

What is the vertices 4y^2-5x^2=20 ?
The vertices 4y^2-5x^2=20 is (0,sqrt(5)),(0,-sqrt(5))