What is the vertices ((y+3)^2)/4-((x+1)^2)/(49)=1 ?

The vertices ((y+3)^2)/4-((x+1)^2)/(49)=1 is (-1,-1),(-1,-5)

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vertices ((y+3)^2)/4-((x+1)^2)/(49)=1

vertices

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{49} = 1

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

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{4} - \frac{( x + 1 ) ^{2}}{49} = 1 :

Up-down Hyperbola with

(h, k) = (- 1, - 3), a = 2, b = 7

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

Refine

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

\frac{( x - 3 ) ^{2}}{16} - \frac{( y - 2 ) ^{2}}{25} = 1

\frac{x ^{2}}{144} - \frac{y ^{2}}{81} = 1

9 y^{2} - 16 x^{2} = 144

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

9 x^{2} - 25 y^{2} - 72 x + 369 = 0

What is the vertices ((y+3)^2)/4-((x+1)^2)/(49)=1 ?
The vertices ((y+3)^2)/4-((x+1)^2)/(49)=1 is (-1,-1),(-1,-5)