What is the vertices (x^2)/(25)-(y^2)/(16)=1 ?

The vertices (x^2)/(25)-(y^2)/(16)=1 is (5,0),(-5,0)

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vertices (x^2)/(25)-(y^2)/(16)=1

vertices

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

(5, 0), (- 5, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 4

(0 + 5, 0), (0 - 5, 0)

Refine

(5, 0), (- 5, 0)

y^{2} + 12 x - 2 y - 11 = 0

x^{2} + 3 y^{2} = 1

x^{2} = 16 y

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

x^{2} + 24 x + y^{2} + 10 y = - 160

What is the vertices (x^2)/(25)-(y^2)/(16)=1 ?
The vertices (x^2)/(25)-(y^2)/(16)=1 is (5,0),(-5,0)