What is the vertices (x^2}{49}-\frac{(y+6)^2)/4 =1 ?

The vertices (x^2}{49}-\frac{(y+6)^2)/4 =1 is (7,-6),(-7,-6)

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

vertices

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

(7, - 6), (- 7, - 6)

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (0, - 6), a = 7, b = 2

(0 + 7, - 6), (0 - 7, - 6)

Refine

(7, - 6), (- 7, - 6)

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

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

\frac{( y ) ^{2}}{40} - \frac{( x ) ^{2}}{81} = 1

\frac{x ^{2}}{25} - \frac{y ^{2}}{121} = 1

6 x^{2} - 8 y^{2} + 12 x - 32 y = 178

What is the vertices (x^2}{49}-\frac{(y+6)^2)/4 =1 ?
The vertices (x^2}{49}-\frac{(y+6)^2)/4 =1 is (7,-6),(-7,-6)