What is the vertices ((y-6)^2)/(36)-((x+1)^2)/(16)=1 ?

The vertices ((y-6)^2)/(36)-((x+1)^2)/(16)=1 is (-1,12),(-1,0)

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vertices ((y-6)^2)/(36)-((x+1)^2)/(16)=1

vertices

\frac{( y - 6 ) ^{2}}{36} - \frac{( x + 1 ) ^{2}}{16} = 1

(- 1, 12), (- 1, 0)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 6 ) ^{2}}{36} - \frac{( x + 1 ) ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (- 1, 6), a = 6, b = 4

(- 1, 6 + 6), (- 1, 6 - 6)

Refine

(- 1, 12), (- 1, 0)

x^{2} - 4 y^{2} - 2 x + 16 y = 20

\frac{( y - 1 ) ^{2}}{16} - \frac{( x + 6 ) ^{2}}{81} = 1

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

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

\frac{x ^{2}}{29} - \frac{y ^{2}}{20} = 1

What is the vertices ((y-6)^2)/(36)-((x+1)^2)/(16)=1 ?
The vertices ((y-6)^2)/(36)-((x+1)^2)/(16)=1 is (-1,12),(-1,0)