What is the foci x^2-9y^2-4x+36y-41=0 ?

The foci x^2-9y^2-4x+36y-41=0 is (2+sqrt(10),2),(2-sqrt(10),2)

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foci x^2-9y^2-4x+36y-41=0

foci

x^{2} - 9 y^{2} - 4 x + 36 y - 41 = 0

(2 + 10, 2), (2 - 10, 2)

Solution steps

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

Calculate Hyperbola properties

x^{2} - 9 y^{2} - 4 x + 36 y - 41 = 0 :

Right-left Hyperbola with

(h, k) = (2, 2), a = 3, b = 1

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

Compute

c :

c = 3^{2} + 1^{2} : 10

(2 + 10, 2), (2 - 10, 2)

4 y^{2} - 9 x^{2} + 40 y - 540 x - 8036 = 0

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

4 x^{2} - 4 y^{2} + 20 x - 16 y + 25 = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{81} = 1

3 x^{2} - 4 y^{2} = 12

What is the foci x^2-9y^2-4x+36y-41=0 ?
The foci x^2-9y^2-4x+36y-41=0 is (2+sqrt(10),2),(2-sqrt(10),2)