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

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

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vertices x^2-4y^2=16

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

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

9 x^{2} - y^{2} - 54 x - 2 y + 71 = 0

\frac{y ^{2}}{25} - \frac{x ^{2}}{36} = 1

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

\frac{( x + 1 ) ^{2}}{9} - \frac{( y - 3 ) ^{2}}{16} = 1

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