What is the foci ((x-3)^2}{25}+\frac{(y-3)^2)/4 =1 ?

The foci ((x-3)^2}{25}+\frac{(y-3)^2)/4 =1 is (3+sqrt(21),3),(3-sqrt(21),3)

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foci ((x-3)^2}{25}+\frac{(y-3)^2)/4 =1

foci

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{4} = 1

(3 + 21, 3), (3 - 21, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (3, 3), a = 5, b = 2

(3 + c, 3), (3 - c, 3)

Compute

c :

c = 5^{2} - 2^{2} : 21

(3 + 21, 3), (3 - 21, 3)

\frac{( x - 3 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{4} = 1

\frac{( x - 8 ) ^{2}}{2} + \frac{( y - 3 ) ^{2}}{27} = 1

x^{2} + 4 y^{2} + 6 x + 16 y + 21 = 0

3 x^{2} + y^{2} = 6

\frac{( x + 1 ) ^{2}}{\frac{3}{2}} + \frac{y ^{2}}{1} = 1

What is the foci ((x-3)^2}{25}+\frac{(y-3)^2)/4 =1 ?
The foci ((x-3)^2}{25}+\frac{(y-3)^2)/4 =1 is (3+sqrt(21),3),(3-sqrt(21),3)