What is the foci ((y+3)^2)/(10)-((x-3)^2)/(20)=1 ?

The foci ((y+3)^2)/(10)-((x-3)^2)/(20)=1 is (3,-3+sqrt(30)),(3,-3-sqrt(30))

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foci ((y+3)^2)/(10)-((x-3)^2)/(20)=1

foci

\frac{( y + 3 ) ^{2}}{10} - \frac{( x - 3 ) ^{2}}{20} = 1

(3, - 3 + 30), (3, - 3 - 30)

Solution steps

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

Calculate Hyperbola properties

\frac{( y + 3 ) ^{2}}{10} - \frac{( x - 3 ) ^{2}}{20} = 1 :

Up-down Hyperbola with

(h, k) = (3, - 3), a = 10, b = 25

(3, - 3 + c), (3, - 3 - c)

Compute

c :

c = 10^{2} + (25)^{2} : 30

(3, - 3 + 30), (3, - 3 - 30)

\frac{x ^{2}}{64} - \frac{y ^{2}}{- 39} = 1

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

1 - \frac{x ^{2}}{4} + (y - 1)^{2} = 0

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

y^{2} - x^{2} = 1

What is the foci ((y+3)^2)/(10)-((x-3)^2)/(20)=1 ?
The foci ((y+3)^2)/(10)-((x-3)^2)/(20)=1 is (3,-3+sqrt(30)),(3,-3-sqrt(30))