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

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

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foci ((x-1)^2)/5+((y-3)^2)/4 =1

foci

\frac{( x - 1 ) ^{2}}{5} + \frac{( y - 3 ) ^{2}}{4} = 1

(2, 3), (0, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{5} + \frac{( y - 3 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (1, 3), a = 5, b = 2

(1 + c, 3), (1 - c, 3)

Compute

c :

c = 5^{2} - 2^{2} : 1

(1 + 1, 3), (1 - 1, 3)

Refine

(2, 3), (0, 3)

\frac{2 x ^{2}}{25} + \frac{( y - 3 ) ^{2}}{49} = 1

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

4 x^{2} + 81 y^{2} = 324

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

7 x^{2} = 21 - 3 y^{2}

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