What is the foci (x^2)/(36)-((y-10)^2)/(64)=1 ?

The foci (x^2)/(36)-((y-10)^2)/(64)=1 is (10,10),(-10,10)

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foci (x^2)/(36)-((y-10)^2)/(64)=1

foci

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1

(10, 10), (- 10, 10)

Solution steps

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

Calculate Hyperbola properties

\frac{x ^{2}}{36} - \frac{( y - 10 ) ^{2}}{64} = 1 :

Right-left Hyperbola with

(h, k) = (0, 10), a = 6, b = 8

(0 + c, 10), (0 - c, 10)

Compute

c :

c = 6^{2} + 8^{2} : 10

(0 + 10, 10), (0 - 10, 10)

Refine

(10, 10), (- 10, 10)

\frac{( y - 1 ) ^{2}}{49} - \frac{( x - 2 ) ^{2}}{32} = 1

\frac{( x - 3 ) ^{2}}{13} - \frac{( y - 5 ) ^{2}}{7} = 1

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

- 4 x^{2} + y^{2} + 6 y - 27 = 0

4 y^{2} - 9 x^{2} + 8 y - 54 x - 113 = 0

What is the foci (x^2)/(36)-((y-10)^2)/(64)=1 ?
The foci (x^2)/(36)-((y-10)^2)/(64)=1 is (10,10),(-10,10)