What is the vertices ((y+3)^2)/9-((x-1)^2)/(18)=1 ?

The vertices ((y+3)^2)/9-((x-1)^2)/(18)=1 is (1,0),(1,-6)

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vertices ((y+3)^2)/9-((x-1)^2)/(18)=1

vertices

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{18} = 1

(1, 0), (1, - 6)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{9} - \frac{( x - 1 ) ^{2}}{18} = 1 :

Up-down Hyperbola with

(h, k) = (1, - 3), a = 3, b = 32

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

Refine

(1, 0), (1, - 6)

\frac{x ^{2}}{1} - \frac{y ^{2}}{35} = 1

\frac{( x + 6 ) ^{2}}{9} = \frac{( y - 2 ) ^{2}}{16} + 1

\frac{x ^{2}}{27} - \frac{y ^{2}}{72} = 1

\frac{y ^{2}}{3} - \frac{x ^{2}}{1} = 1

4 x^{2} - 5 y^{2} + 30 y = 65

What is the vertices ((y+3)^2)/9-((x-1)^2)/(18)=1 ?
The vertices ((y+3)^2)/9-((x-1)^2)/(18)=1 is (1,0),(1,-6)