What is the vertices (y^2)/4-(x^2)/(21)=1 ?

The vertices (y^2)/4-(x^2)/(21)=1 is (0,2),(0,-2)

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vertices (y^2)/4-(x^2)/(21)=1

vertices

\frac{y ^{2}}{4} - \frac{x ^{2}}{21} = 1

(0, 2), (0, - 2)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{4} - \frac{x ^{2}}{21} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 2, b = 21

(0, 0 + 2), (0, 0 - 2)

Refine

(0, 2), (0, - 2)

\frac{x ^{2}}{49} - \frac{y ^{2}}{81} = 1

\frac{x ^{2}}{49} - \frac{y ^{2}}{36} = 1

\frac{y ^{2}}{25} - \frac{x ^{2}}{64} = 1

9 y^{2} - 25 x^{2} = 225

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

What is the vertices (y^2)/4-(x^2)/(21)=1 ?
The vertices (y^2)/4-(x^2)/(21)=1 is (0,2),(0,-2)