What is the vertices x^2-4y^2-2x+16y=20 ?

The vertices x^2-4y^2-2x+16y=20 is (1+sqrt(5),2),(1-sqrt(5),2)

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vertices x^2-4y^2-2x+16y=20

vertices

x^{2} - 4 y^{2} - 2 x + 16 y = 20

(1 + 5, 2), (1 - 5, 2)

Solution steps

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

Calculate Hyperbola properties

x^{2} - 4 y^{2} - 2 x + 16 y = 20 :

Right-left Hyperbola with

(h, k) = (1, 2), a = 5, b = \frac{5}{2}

(1 + 5, 2), (1 - 5, 2)

\frac{( y - 1 ) ^{2}}{16} - \frac{( x + 6 ) ^{2}}{81} = 1

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

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

\frac{x ^{2}}{29} - \frac{y ^{2}}{20} = 1

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

What is the vertices x^2-4y^2-2x+16y=20 ?
The vertices x^2-4y^2-2x+16y=20 is (1+sqrt(5),2),(1-sqrt(5),2)