What is the vertices+(x^2)/4-(y^2)/9 =1 ?

The vertices+(x^2)/4-(y^2)/9 =1 is (2,0),(-2,0)

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vertices+(x^2)/4-(y^2)/9 =1

vertices

+ \frac{x ^{2}}{4} - \frac{y ^{2}}{9} = 1

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

Solution steps

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

Calculate Hyperbola properties

+ \frac{x ^{2}}{4} - \frac{y ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 2, b = 3

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

Refine

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

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

5 x^{2} - 5 y^{2} + 20 x - 60 y = 185

9 x^{2} - 4 y^{2} + 54 x - 8 y + 78 = 0

x^{2} - y^{2} < 1

x^{2} - y^{2} = 9

What is the vertices+(x^2)/4-(y^2)/9 =1 ?
The vertices+(x^2)/4-(y^2)/9 =1 is (2,0),(-2,0)