What is the vertices 9x^2-16y^2-18x+64y-199=0 ?

The vertices 9x^2-16y^2-18x+64y-199=0 is (5,2),(-3,2)

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vertices 9x^2-16y^2-18x+64y-199=0

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

(h, k) = (1, 2), a = 4, b = 3

(1 + 4, 2), (1 - 4, 2)

Refine

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

\frac{x ^{2}}{25} - \frac{y ^{2}}{49} = 1

- 2 x^{2} + 5 y^{2} + 3 x + 8 y + 4 = 0

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

(4 x - 3 y) = \frac{- 144}{( 4 x + 3 y )}

4 y^{2} - 5 x^{2} = 20

What is the vertices 9x^2-16y^2-18x+64y-199=0 ?
The vertices 9x^2-16y^2-18x+64y-199=0 is (5,2),(-3,2)