What is the vertices (x^2)/(16)-(y^2)/(16)=1 ?

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

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vertices (x^2)/(16)-(y^2)/(16)=1

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

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

\frac{y ^{2}}{36} - \frac{x ^{2}}{81} = 1

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

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

\frac{( y - 8 ) ^{2}}{4} - \frac{( x - 5 ) ^{2}}{9} = 1

7 x^{2} - 9 y^{2} - 14 x + 72 y - 200 = 0

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