What is the vertices ((y+2)^2)/4-((x-1)^2)/(45)=1 ?

The vertices ((y+2)^2)/4-((x-1)^2)/(45)=1 is (1,0),(1,-4)

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vertices ((y+2)^2)/4-((x-1)^2)/(45)=1

vertices

\frac{( y + 2 ) ^{2}}{4} - \frac{( x - 1 ) ^{2}}{45} = 1

(1, 0), (1, - 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 2 ) ^{2}}{4} - \frac{( x - 1 ) ^{2}}{45} = 1 :

Up-down Hyperbola with

(h, k) = (1, - 2), a = 2, b = 35

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

Refine

(1, 0), (1, - 4)

\frac{x ^{2}}{729} - \frac{y ^{2}}{484} = 1

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

\frac{( x + 3 ) ^{2}}{49} - \frac{( y - 1 ) ^{2}}{48} = 1

\frac{9}{125} y^{2} - \frac{1}{5} x^{2} = 1

y^{2} - \frac{x ^{2}}{4} = 1

What is the vertices ((y+2)^2)/4-((x-1)^2)/(45)=1 ?
The vertices ((y+2)^2)/4-((x-1)^2)/(45)=1 is (1,0),(1,-4)