What is the foci ((x-5)^2)/4+((y+3)^2)/(16)=1 ?

The foci ((x-5)^2)/4+((y+3)^2)/(16)=1 is (5,-3+2sqrt(3)),(5,-3-2sqrt(3))

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foci ((x-5)^2)/4+((y+3)^2)/(16)=1

foci

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

(5, - 3 + 23), (5, - 3 - 23)

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (5, - 3), b = 4, a = 2

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

Compute

c :

c = 4^{2} - 2^{2} : 23

(5, - 3 + 23), (5, - 3 - 23)

(x - 2)^{2} = 8 y

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

\frac{x ^{2}}{36} + \frac{y ^{2}}{9} = 1

x^{2} + y^{2} - 14 x + 2 y + 41 = 0

y = x^{2} + 3

What is the foci ((x-5)^2)/4+((y+3)^2)/(16)=1 ?
The foci ((x-5)^2)/4+((y+3)^2)/(16)=1 is (5,-3+2sqrt(3)),(5,-3-2sqrt(3))