What is the vertices (y^2}{16}-\frac{x^2)/4 =1 ?

The vertices (y^2}{16}-\frac{x^2)/4 =1 is (0,4),(0,-4)

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vertices (y^2}{16}-\frac{x^2)/4 =1

vertices

\frac{y ^{2}}{16} - \frac{x ^{2}}{4} = 1

(0, 4), (0, - 4)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{16} - \frac{x ^{2}}{4} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 4, b = 2

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

Refine

(0, 4), (0, - 4)

x^{2} - y^{2} - 4 x + 4 y + 9 = 0

x^{2} - y^{2} + 8 x - 2 y - 10 = 0

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

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

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

What is the vertices (y^2}{16}-\frac{x^2)/4 =1 ?
The vertices (y^2}{16}-\frac{x^2)/4 =1 is (0,4),(0,-4)