What is the vertices ((y-3)^2)/(49)-((x+2)^2)/(16)=1 ?

The vertices ((y-3)^2)/(49)-((x+2)^2)/(16)=1 is (-2,10),(-2,-4)

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vertices ((y-3)^2)/(49)-((x+2)^2)/(16)=1

vertices

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1

(- 2, 10), (- 2, - 4)

Solution steps

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

Calculate Hyperbola properties

\frac{( y - 3 ) ^{2}}{49} - \frac{( x + 2 ) ^{2}}{16} = 1 :

Up-down Hyperbola with

(h, k) = (- 2, 3), a = 7, b = 4

(- 2, 3 + 7), (- 2, 3 - 7)

Refine

(- 2, 10), (- 2, - 4)

x^{2} - y^{2} - 4 x + 8 y - 21 = 0

16 y^{2} - 32 y - 9 x^{2} - 18 x = 137

10 x^{2} - 25 y^{2} = 100

\frac{y ^{2}}{64} - \frac{x ^{2}}{36} = 1

4 x^{2} - y^{2} - 24 x - 4 y + 28 = 0

What is the vertices ((y-3)^2)/(49)-((x+2)^2)/(16)=1 ?
The vertices ((y-3)^2)/(49)-((x+2)^2)/(16)=1 is (-2,10),(-2,-4)