What is the vertices 10y^2-5x^2=50 ?

The vertices 10y^2-5x^2=50 is (0,sqrt(5)),(0,-sqrt(5))

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vertices 10y^2-5x^2=50

vertices

10 y^{2} - 5 x^{2} = 50

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

Solution steps

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

Calculate Hyperbola properties

10 y^{2} - 5 x^{2} = 50 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 5, b = 10

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

Refine

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

\frac{x ^{2}}{9} - \frac{y ^{2}}{36} = 1

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

2 x^{2} - 2 y^{2} - 20 x + 16 y - 16 = 0

- 9 x^{2} + 4 y^{2} = 36

\frac{( x - 1 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{16} = 1

What is the vertices 10y^2-5x^2=50 ?
The vertices 10y^2-5x^2=50 is (0,sqrt(5)),(0,-sqrt(5))