What is the vertices ((y+3)^2)/(25)-((x-2)^2)/(49)=1 ?

The vertices ((y+3)^2)/(25)-((x-2)^2)/(49)=1 is (2,2),(2,-8)

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vertices ((y+3)^2)/(25)-((x-2)^2)/(49)=1

vertices

\frac{( y + 3 ) ^{2}}{25} - \frac{( x - 2 ) ^{2}}{49} = 1

(2, 2), (2, - 8)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{25} - \frac{( x - 2 ) ^{2}}{49} = 1 :

Up-down Hyperbola with

(h, k) = (2, - 3), a = 5, b = 7

(2, - 3 + 5), (2, - 3 - 5)

Refine

(2, 2), (2, - 8)

\frac{y ^{2}}{15} - \frac{x ^{2}}{1} = 1

x^{2} - 4 y^{2} + 8 x + 16 y - 4 = 0

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

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

9 (x - 1)^{2} - 16 (y + 2)^{2} = 144

What is the vertices ((y+3)^2)/(25)-((x-2)^2)/(49)=1 ?
The vertices ((y+3)^2)/(25)-((x-2)^2)/(49)=1 is (2,2),(2,-8)