What is the foci ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1 ?

The foci ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1 is (4+3sqrt(5),2),(4-3sqrt(5),2)

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foci ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1

foci

\frac{( x - 4 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{9} = 1

(4 + 35, 2), (4 - 35, 2)

Solution steps

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

Calculate Hyperbola properties

\frac{( x - 4 ) ^{2}}{36} - \frac{( y - 2 ) ^{2}}{9} = 1 :

Right-left Hyperbola with

(h, k) = (4, 2), a = 6, b = 3

(4 + c, 2), (4 - c, 2)

Compute

c :

c = 6^{2} + 3^{2} : 35

(4 + 35, 2), (4 - 35, 2)

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

x = (y - 5)^{2}

(x - 2)^{2} = 12 (y + 3)

(y - 2)^{2} = 12 (x + 1)

f (x) = x^{2} + 2 x

What is the foci ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1 ?
The foci ((x-4)^2}{36}-\frac{(y-2)^2)/9 =1 is (4+3sqrt(5),2),(4-3sqrt(5),2)