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

The vertices ((y-1)^2)/9-((x-2)^2)/(16)=1 is (2,4),(2,-2)

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

vertices

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

(2, 4), (2, - 2)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

(2, 4), (2, - 2)

\frac{( x + 1 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{64} = 1

\frac{( x + 3 ) ^{2}}{16} - \frac{( y + 2 ) ^{2}}{36} = 1

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

\frac{x ^{2}}{25} - \frac{y ^{2}}{81} = 1

\frac{( x - 3 ) ^{2}}{9} - \frac{( y + 4 ) ^{2}}{4} = 1

What is the vertices ((y-1)^2)/9-((x-2)^2)/(16)=1 ?
The vertices ((y-1)^2)/9-((x-2)^2)/(16)=1 is (2,4),(2,-2)