What is the foci x^2+4y^2-2x-16y+13=0 ?

The foci x^2+4y^2-2x-16y+13=0 is (1+sqrt(3),2),(1-sqrt(3),2)

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foci x^2+4y^2-2x-16y+13=0

foci

x^{2} + 4 y^{2} - 2 x - 16 y + 13 = 0

(1 + 3, 2), (1 - 3, 2)

Solution steps

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

Calculate ellipse properties

x^{2} + 4 y^{2} - 2 x - 16 y + 13 = 0 :

Ellipse with center

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

(1 + c, 2), (1 - c, 2)

Compute

c :

c = 2^{2} - 1^{2} : 3

(1 + 3, 2), (1 - 3, 2)

4 x^{2} + 9 y^{2} = 25

4 y^{2} + 2 x^{2} - 4 y - 8 x - 15 = 0

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

\frac{( x - 4 ) ^{2}}{9} + \frac{( y - 7 ) ^{2}}{36} = 1

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

What is the foci x^2+4y^2-2x-16y+13=0 ?
The foci x^2+4y^2-2x-16y+13=0 is (1+sqrt(3),2),(1-sqrt(3),2)