What is the foci ((x-2)^2}{36}+\frac{(y-6)^2)/4 =1 ?

The foci ((x-2)^2}{36}+\frac{(y-6)^2)/4 =1 is (2+4sqrt(2),6),(2-4sqrt(2),6)

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foci ((x-2)^2}{36}+\frac{(y-6)^2)/4 =1

foci

\frac{( x - 2 ) ^{2}}{36} + \frac{( y - 6 ) ^{2}}{4} = 1

(2 + 42, 6), (2 - 42, 6)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{36} + \frac{( y - 6 ) ^{2}}{4} = 1 :

Ellipse with center

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

(2 + c, 6), (2 - c, 6)

Compute

c :

c = 6^{2} - 2^{2} : 42

(2 + 42, 6), (2 - 42, 6)

4 x^{2} + y^{2} - 48 x + 4 y + 48 = 0

7 x^{2} = 14 - 2 y^{2}

25 x^{2} + 4 y^{2} - 100 = 0

\frac{x ^{2}}{1.586 9 ^{2}} + \frac{y ^{2}}{0.9989} = 1

\frac{x ^{2}}{36} + \frac{y ^{2}}{49} = 1

What is the foci ((x-2)^2}{36}+\frac{(y-6)^2)/4 =1 ?
The foci ((x-2)^2}{36}+\frac{(y-6)^2)/4 =1 is (2+4sqrt(2),6),(2-4sqrt(2),6)