What is the vertices ((y+3)^2)/9-((x-2)^2)/(16)=1 ?

The vertices ((y+3)^2)/9-((x-2)^2)/(16)=1 is (2,0),(2,-6)

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vertices ((y+3)^2)/9-((x-2)^2)/(16)=1

vertices

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1

(2, 0), (2, - 6)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1 :

Up-down Hyperbola with

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

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

Refine

(2, 0), (2, - 6)

9 x^{2} - 18 x - 24 y - 63 = 4 y^{2}

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

\frac{( x - 2 ) ^{2}}{4} - \frac{( y + 3 ) ^{2}}{5} = 1

\frac{( x - 8 ) ^{2}}{9} - \frac{( y + 3 ) ^{2}}{16} = 1

25 y^{2} - x^{2} = 100

What is the vertices ((y+3)^2)/9-((x-2)^2)/(16)=1 ?
The vertices ((y+3)^2)/9-((x-2)^2)/(16)=1 is (2,0),(2,-6)