What is the foci 3x^2+12y^2-6x-48y+33=0 ?

The foci 3x^2+12y^2-6x-48y+33=0 is ((sqrt(2)+3)/(sqrt(2)),2),((sqrt(2)-3)/(sqrt(2)),2)

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foci 3x^2+12y^2-6x-48y+33=0

foci

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0

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

Solution steps

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

Calculate ellipse properties

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0 :

Ellipse with center

(h, k) = (1, 2), a = 6, b = \frac{3}{2}

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

Compute

c :

c = 6^{2} - \frac{3}{2}^{2} : \frac{3}{2}

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

Refine

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

16 x^{2} - 64 x + 9 y^{2} = 80

4 x^{2} + 9 y^{2} - 8 x - 32 = 0

\frac{( x - 4 ) ^{2}}{32} + \frac{( y - 3 ) ^{2}}{36} = 1

\frac{( x + 3 ) ^{2}}{21} + \frac{( y - 6 ) ^{2}}{25} = 1

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

What is the foci 3x^2+12y^2-6x-48y+33=0 ?
The foci 3x^2+12y^2-6x-48y+33=0 is ((sqrt(2)+3)/(sqrt(2)),2),((sqrt(2)-3)/(sqrt(2)),2)