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

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

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

vertices

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

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

Solution steps

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

Calculate Hyperbola properties

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

Right-left Hyperbola with

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

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

Refine

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

\frac{y ^{2}}{4 ^{2}} - \frac{x ^{2}}{3 ^{2}} = 1

\frac{( y + 2 ) ^{2}}{4} - \frac{( x - 1 ) ^{2}}{45} = 1

\frac{x ^{2}}{729} - \frac{y ^{2}}{484} = 1

\frac{x ^{2}}{25} - \frac{y ^{2}}{144} = 1

\frac{( x + 3 ) ^{2}}{49} - \frac{( y - 1 ) ^{2}}{48} = 1

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