What is the vertices (x^2}{49}-\frac{y^2)/4 =1 ?

The vertices (x^2}{49}-\frac{y^2)/4 =1 is (7,0),(-7,0)

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vertices (x^2}{49}-\frac{y^2)/4 =1

vertices

\frac{x ^{2}}{49} - \frac{y ^{2}}{4} = 1

(7, 0), (- 7, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{49} - \frac{y ^{2}}{4} = 1 :

Right-left Hyperbola with

(h, k) = (0, 0), a = 7, b = 2

(0 + 7, 0), (0 - 7, 0)

Refine

(7, 0), (- 7, 0)

y^{2} - 25 x^{2} + 8 y - 9 = 0

x^{2} - 16 y^{2} = 144

25 x^{2} - 16 y^{2} + 50 x - 32 y - 391 = 0

\frac{x ^{2}}{49} - \frac{( y + 6 ) ^{2}}{4} = 1

5 y^{2} - 16 x^{2} + 400 = 0

What is the vertices (x^2}{49}-\frac{y^2)/4 =1 ?
The vertices (x^2}{49}-\frac{y^2)/4 =1 is (7,0),(-7,0)