What is the vertices 4x^2-25y^2+16x+50y-109=0 ?

The vertices 4x^2-25y^2+16x+50y-109=0 is (3,1),(-7,1)

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vertices 4x^2-25y^2+16x+50y-109=0

vertices

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0

(3, 1), (- 7, 1)

Solution steps

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

Calculate Hyperbola properties

4 x^{2} - 25 y^{2} + 16 x + 50 y - 109 = 0 :

Right-left Hyperbola with

(h, k) = (- 2, 1), a = 5, b = 2

(- 2 + 5, 1), (- 2 - 5, 1)

Refine

(3, 1), (- 7, 1)

\frac{y ^{2}}{16} - \frac{x ^{2}}{4} = 1

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

x^{2} - y^{2} + 8 x - 2 y - 10 = 0

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

9 x^{2} - y^{2} = 36

What is the vertices 4x^2-25y^2+16x+50y-109=0 ?
The vertices 4x^2-25y^2+16x+50y-109=0 is (3,1),(-7,1)