What is the vertices 16x^2-9y^2+64x-90y=305 ?

The vertices 16x^2-9y^2+64x-90y=305 is (1,-5),(-5,-5)

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vertices 16x^2-9y^2+64x-90y=305

vertices

16 x^{2} - 9 y^{2} + 64 x - 90 y = 305

(1, - 5), (- 5, - 5)

Solution steps

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

Calculate Hyperbola properties

16 x^{2} - 9 y^{2} + 64 x - 90 y = 305 :

Right-left Hyperbola with

(h, k) = (- 2, - 5), a = 3, b = 4

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

Refine

(1, - 5), (- 5, - 5)

4 x^{2} - 9 y^{2} - 36 < 0

\frac{x ^{2}}{16} - \frac{y ^{2}}{64} = 1

x^{2} - 3 y^{2} + 2 x + 6 y - 1 = 0

9 x^{2} - 16 y^{2} - 18 x - 64 y - 199 = 0

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

What is the vertices 16x^2-9y^2+64x-90y=305 ?
The vertices 16x^2-9y^2+64x-90y=305 is (1,-5),(-5,-5)