What is the vertices ((y-3)^2)/(16)-((x+4)^2)/(25)=1 ?

The vertices ((y-3)^2)/(16)-((x+4)^2)/(25)=1 is (-4,7),(-4,-1)

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vertices ((y-3)^2)/(16)-((x+4)^2)/(25)=1

vertices

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1

(- 4, 7), (- 4, - 1)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 3 ) ^{2}}{16} - \frac{( x + 4 ) ^{2}}{25} = 1 :

Up-down Hyperbola with

(h, k) = (- 4, 3), a = 4, b = 5

(- 4, 3 + 4), (- 4, 3 - 4)

Refine

(- 4, 7), (- 4, - 1)

x^{2} - y^{2} + 2 y = 2

9 y^{2} + 36 y - 4 x^{2} - 32 x - 64 = 0

- 25 x^{2} + 9 y^{2} = 225

\frac{( y - 2 ) ^{2}}{9} - \frac{( x + 1 ) ^{2}}{25} = 1

9 y^{2} - x^{2} + 54 y + 2 x + 62 = 0

What is the vertices ((y-3)^2)/(16)-((x+4)^2)/(25)=1 ?
The vertices ((y-3)^2)/(16)-((x+4)^2)/(25)=1 is (-4,7),(-4,-1)