What is the vertices 4y^2-49x^2-392x+16y-964=0 ?

The vertices 4y^2-49x^2-392x+16y-964=0 is (-4,5),(-4,-9)

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vertices 4y^2-49x^2-392x+16y-964=0

vertices

4 y^{2} - 49 x^{2} - 392 x + 16 y - 964 = 0

(- 4, 5), (- 4, - 9)

Solution steps

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

Calculate Hyperbola properties

4 y^{2} - 49 x^{2} - 392 x + 16 y - 964 = 0 :

Up-down Hyperbola with

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

(- 4, - 2 + 7), (- 4, - 2 - 7)

Refine

(- 4, 5), (- 4, - 9)

5 x^{2} - 2 y^{2} + 10 x - 12 y = 23

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

3 x^{2} - 15 y^{2} + 2 x - 3 y = 4

\frac{y ^{2}}{4} - \frac{x ^{2}}{9} = 1

\frac{y ^{2}}{16} - \frac{x ^{2}}{36} = 1

What is the vertices 4y^2-49x^2-392x+16y-964=0 ?
The vertices 4y^2-49x^2-392x+16y-964=0 is (-4,5),(-4,-9)