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

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

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

foci

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

(1, - 2 + 25), (1, - 2 - 25)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

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

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

Compute

c :

c = 2^{2} + 4^{2} : 25

(1, - 2 + 25), (1, - 2 - 25)

x^{2} - 2 x - 3 = y^{2} + 2 y + 3

6 x^{2} - 5 y^{2} - 24 x + 10 y - 11 = 0

4 x^{2} - 24 x - 25 y^{2} + 250 y - 489 = 0

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

9 y^{2} - 4 x^{2} + 32 x - 36 y - 64 = 0

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