What is the foci 4x^2+2y^2+2x-8y+7=0 ?

The foci 4x^2+2y^2+2x-8y+7=0 is (-1/4 ,(8+sqrt(5))/4),(-1/4 ,(8-sqrt(5))/4)

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foci 4x^2+2y^2+2x-8y+7=0

foci

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0

(- \frac{1}{4}, \frac{8 + 5}{4}), (- \frac{1}{4}, \frac{8 - 5}{4})

Solution steps

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

Calculate ellipse properties

4 x^{2} + 2 y^{2} + 2 x - 8 y + 7 = 0 :

Ellipse with center

(h, k) = (- \frac{1}{4}, 2), b = \frac{5}{2 2}, a = \frac{5}{4}

(- \frac{1}{4}, 2 + c), (- \frac{1}{4}, 2 - c)

Compute

c :

c = (\frac{5}{2 2})^{2} - (\frac{5}{4})^{2} : \frac{5}{4}

(- \frac{1}{4}, 2 + \frac{5}{4}), (- \frac{1}{4}, 2 - \frac{5}{4})

Refine

(- \frac{1}{4}, \frac{8 + 5}{4}), (- \frac{1}{4}, \frac{8 - 5}{4})

9 x^{2} - 36 x + 4 y^{2} = 0

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

25 x^{2} + 16 y^{2} + 50 x - 64 y - 311 = 0

x^{2} + 8 (y + 5)^{2} = 20

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} > 1

What is the foci 4x^2+2y^2+2x-8y+7=0 ?
The foci 4x^2+2y^2+2x-8y+7=0 is (-1/4 ,(8+sqrt(5))/4),(-1/4 ,(8-sqrt(5))/4)