What is the vertices (x^2)/(25)-(y^2)/(144)=1 ?

The vertices (x^2)/(25)-(y^2)/(144)=1 is (5,0),(-5,0)

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vertices (x^2)/(25)-(y^2)/(144)=1

vertices

\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 5, b = 12

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

Refine

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

\frac{( x + 3 ) ^{2}}{49} - \frac{( y - 1 ) ^{2}}{48} = 1

\frac{9}{125} y^{2} - \frac{1}{5} x^{2} = 1

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

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

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

What is the vertices (x^2)/(25)-(y^2)/(144)=1 ?
The vertices (x^2)/(25)-(y^2)/(144)=1 is (5,0),(-5,0)