What is the foci ((y)^2)/(40)-((x)^2)/(81)=1 ?

The foci ((y)^2)/(40)-((x)^2)/(81)=1 is (0,11),(0,-11)

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foci ((y)^2)/(40)-((x)^2)/(81)=1

foci

\frac{( y ) ^{2}}{40} - \frac{( x ) ^{2}}{81} = 1

(0, 11), (0, - 11)

Solution steps

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

Calculate Hyperbola properties

\frac{y ^{2}}{40} - \frac{x ^{2}}{81} = 1 :

Up-down Hyperbola with

(h, k) = (0, 0), a = 210, b = 9

(0, 0 + c), (0, 0 - c)

Compute

c :

c = (210)^{2} + 9^{2} : 11

(0, 0 + 11), (0, 0 - 11)

Refine

(0, 11), (0, - 11)

+ 9 x^{2} - 16 y^{2} = 144

\frac{x ^{2}}{4} - \frac{( y + 3 ) ^{2}}{9} = 1

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

\frac{( x + 5 ) ^{2}}{36} - \frac{( y + 1 ) ^{2}}{64} = 1

x^{2} - y^{2} - 4 x - 2 y - 1 = 0

What is the foci ((y)^2)/(40)-((x)^2)/(81)=1 ?
The foci ((y)^2)/(40)-((x)^2)/(81)=1 is (0,11),(0,-11)