What is the foci ((x-5)^2)/4+((y-3)^2)/(49)=1 ?

The foci ((x-5)^2)/4+((y-3)^2)/(49)=1 is (5,3+3sqrt(5)),(5,3-3sqrt(5))

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foci ((x-5)^2)/4+((y-3)^2)/(49)=1

foci

\frac{( x - 5 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{49} = 1

(5, 3 + 35), (5, 3 - 35)

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (5, 3), b = 7, a = 2

(5, 3 + c), (5, 3 - c)

Compute

c :

c = 7^{2} - 2^{2} : 35

(5, 3 + 35), (5, 3 - 35)

\frac{( x + 5 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{25} = 1

4 x^{2} + 9 y^{2} + 16 x + 18 y - 11 = 0

\frac{x ^{2}}{144} + \frac{y ^{2}}{225} = 1

4 x^{2} + 25 y^{2} + 40 x - 150 y + 225 = 0

\frac{x ^{2}}{\frac{81}{4}} + \frac{y ^{2}}{\frac{25}{16}} = 1

What is the foci ((x-5)^2)/4+((y-3)^2)/(49)=1 ?
The foci ((x-5)^2)/4+((y-3)^2)/(49)=1 is (5,3+3sqrt(5)),(5,3-3sqrt(5))