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

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

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Refine

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

\frac{x ^{2}}{36} - \frac{y ^{2}}{81} = 1

\frac{( y - 1 ) ^{2}}{9} - \frac{( x - 2 ) ^{2}}{16} = 1

\frac{( x + 1 ) ^{2}}{25} - \frac{( y + 4 ) ^{2}}{64} = 1

\frac{( x + 3 ) ^{2}}{16} - \frac{( y + 2 ) ^{2}}{36} = 1

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

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