What is the foci ((x-2)^2)/9+((y-3)^2)/5 =1 ?

The foci ((x-2)^2)/9+((y-3)^2)/5 =1 is (4,3),(0,3)

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foci ((x-2)^2)/9+((y-3)^2)/5 =1

foci

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1

(4, 3), (0, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{5} = 1 :

Ellipse with center

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

(2 + c, 3), (2 - c, 3)

Compute

c :

c = 3^{2} - 5^{2} : 2

(2 + 2, 3), (2 - 2, 3)

Refine

(4, 3), (0, 3)

\frac{( x - 3 ) ^{2}}{100} + \frac{( y - 1 ) ^{2}}{64} = 1

25 x^{2} + y^{2} - 4 y - 21 = 0

16 x^{2} + 25 y^{2} - 64 x + 150 y = - 279

\frac{( x - 12 ) ^{2}}{100} + \frac{( y + 7 ) ^{2}}{36} = 1

\frac{x ^{2}}{0.38 4 ^{2}} + \frac{y ^{2}}{0.375 8 ^{2}} = 1

What is the foci ((x-2)^2)/9+((y-3)^2)/5 =1 ?
The foci ((x-2)^2)/9+((y-3)^2)/5 =1 is (4,3),(0,3)