What is the vertices 4x^2-y^2=16 ?

The vertices 4x^2-y^2=16 is (2,0),(-2,0)

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vertices 4x^2-y^2=16

vertices

4 x^{2} - y^{2} = 16

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

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - y^{2} = 16 :

Right-left Hyperbola with

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

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

Refine

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

144 y^{2} - 25 x^{2} = 3600

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

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

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

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