What is the vertices ((y-2)^2)/9-((x-1)^2)/(16)=1 ?

The vertices ((y-2)^2)/9-((x-1)^2)/(16)=1 is (1,5),(1,-1)

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vertices ((y-2)^2)/9-((x-1)^2)/(16)=1

vertices

\frac{( y - 2 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{16} = 1

(1, 5), (1, - 1)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 2 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (1, 2), a = 3, b = 4

(1, 2 + 3), (1, 2 - 3)

Refine

(1, 5), (1, - 1)

\frac{x ^{2}}{16} + \frac{y ^{2}}{25} = 1

y > x^{2} - 3

\frac{x ^{2}}{36} - \frac{y ^{2}}{64} = 1

x^{2} + \frac{y ^{2}}{16} = 1

(x + 1)^{2} + (y - 2)^{2} = 9

What is the vertices ((y-2)^2)/9-((x-1)^2)/(16)=1 ?
The vertices ((y-2)^2)/9-((x-1)^2)/(16)=1 is (1,5),(1,-1)