What is the foci 4x^2+16y^2-4x-32y+1=0 ?

The foci 4x^2+16y^2-4x-32y+1=0 is ((1+2sqrt(3))/2 ,1),((1-2sqrt(3))/2 ,1)

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foci 4x^2+16y^2-4x-32y+1=0

foci

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0

(\frac{1 + 2 3}{2}, 1), (\frac{1 - 2 3}{2}, 1)

Solution steps

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

Calculate ellipse properties

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0 :

Ellipse with center

(h, k) = (\frac{1}{2}, 1), a = 2, b = 1

(\frac{1}{2} + c, 1), (\frac{1}{2} - c, 1)

Compute

c :

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

(\frac{1}{2} + 3, 1), (\frac{1}{2} - 3, 1)

Refine

(\frac{1 + 2 3}{2}, 1), (\frac{1 - 2 3}{2}, 1)

\frac{( x - 1 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{5100} = 1

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1

(x + 4)^{2} + 9 (y - 3)^{2} = 81

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

4 x^{2} + 25 y^{2} - 50 y - 75 = 0

What is the foci 4x^2+16y^2-4x-32y+1=0 ?
The foci 4x^2+16y^2-4x-32y+1=0 is ((1+2sqrt(3))/2 ,1),((1-2sqrt(3))/2 ,1)