What is the vertices 25x^2-16y^2+50x+128y-631=0 ?

The vertices 25x^2-16y^2+50x+128y-631=0 is (3,4),(-5,4)

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vertices 25x^2-16y^2+50x+128y-631=0

vertices

25 x^{2} - 16 y^{2} + 50 x + 128 y - 631 = 0

(3, 4), (- 5, 4)

Solution steps

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

Calculate Hyperbola properties

25 x^{2} - 16 y^{2} + 50 x + 128 y - 631 = 0 :

Right-left Hyperbola with

(h, k) = (- 1, 4), a = 4, b = 5

(- 1 + 4, 4), (- 1 - 4, 4)

Refine

(3, 4), (- 5, 4)

x^{2} - y^{2} + 7 x + 8 y = 5

16 x^{2} - 25 y^{2} = 400

x^{2} - 9 y^{2} - 4 x + 36 y - 41 = 0

\frac{x ^{2}}{9} - \frac{y ^{2}}{49} = 1

4 x^{2} - \frac{1}{2} y^{2} + 8 x + 2 y = 2

What is the vertices 25x^2-16y^2+50x+128y-631=0 ?
The vertices 25x^2-16y^2+50x+128y-631=0 is (3,4),(-5,4)