What is the vertices 16y^2-x^2+2x+64y+47=0 ?

The vertices 16y^2-x^2+2x+64y+47=0 is (1,-1),(1,-3)

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vertices 16y^2-x^2+2x+64y+47=0

vertices

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0

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

Solution steps

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

Calculate Hyperbola properties

16 y^{2} - x^{2} + 2 x + 64 y + 47 = 0 :

Up-down Hyperbola with

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

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

Refine

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

\frac{x ^{2}}{45} - \frac{y ^{2}}{360} = 1

49 x^{2} - 25 y^{2} - 686 x - 50 y + 2375 = 0

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

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

\frac{( x - 6 ) ^{2}}{48} - (y - 2)^{2} = 1

What is the vertices 16y^2-x^2+2x+64y+47=0 ?
The vertices 16y^2-x^2+2x+64y+47=0 is (1,-1),(1,-3)