What is the foci ((y-1)^2)/(49)-((x-2)^2)/(32)=1 ?

The foci ((y-1)^2)/(49)-((x-2)^2)/(32)=1 is (2,10),(2,-8)

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foci ((y-1)^2)/(49)-((x-2)^2)/(32)=1

foci

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

(2, 10), (2, - 8)

Solution steps

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

Calculate Hyperbola properties

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

Up-down Hyperbola with

(h, k) = (2, 1), a = 7, b = 42

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

Compute

c :

c = 7^{2} + (42)^{2} : 9

(2, 1 + 9), (2, 1 - 9)

Refine

(2, 10), (2, - 8)

\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

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

What is the foci ((y-1)^2)/(49)-((x-2)^2)/(32)=1 ?
The foci ((y-1)^2)/(49)-((x-2)^2)/(32)=1 is (2,10),(2,-8)