What is the foci ((y-1)^2)/(1/4)-((x+1)^2)/(1/16)=1 ?

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

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foci ((y-1)^2)/(1/4)-((x+1)^2)/(1/16)=1

foci

\frac{( y - 1 ) ^{2}}{\frac{1}{4}} - \frac{( x + 1 ) ^{2}}{\frac{1}{16}} = 1

(- 1, \frac{4 + 5}{4}), (- 1, \frac{4 - 5}{4})

Solution steps

(h, k + c), (h, k - c)

Calculate Hyperbola properties

\frac{( y - 1 ) ^{2}}{\frac{1}{4}} - \frac{( x + 1 ) ^{2}}{\frac{1}{16}} = 1 :

Up-down Hyperbola with

(h, k) = (- 1, 1), a = \frac{1}{2}, b = \frac{1}{4}

(- 1, 1 + c), (- 1, 1 - c)

Compute

c :

c = (\frac{1}{2})^{2} + (\frac{1}{4})^{2} : \frac{5}{4}

(- 1, 1 + \frac{5}{4}), (- 1, 1 - \frac{5}{4})

Refine

(- 1, \frac{4 + 5}{4}), (- 1, \frac{4 - 5}{4})

64 y^{2} - 4 x^{2} = 256

4 x^{2} - y^{2} - 24 x + 20 = 0

\frac{( x + 3 ) ^{2}}{9} - \frac{( y - 2 ) ^{2}}{16} = 1

9 x^{2} - y^{2} - 54 x - 2 y - 1 = 0

\frac{x ^{2}}{60} - \frac{y ^{2}}{40} = 1

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