What is the foci 4x^2+y^2+40x+6y+95=0 ?

The foci 4x^2+y^2+40x+6y+95=0 is (-5,-3+sqrt(21/2)),(-5,-3-sqrt(21/2))

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foci 4x^2+y^2+40x+6y+95=0

foci

(- 5, - 3 + \frac{21}{2}), (- 5, - 3 - \frac{21}{2})

Solution steps

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

Calculate ellipse properties

4 x^{2} + y^{2} + 40 x + 6 y + 95 = 0 :

Ellipse with center

(h, k) = (- 5, - 3), b = 14, a = \frac{7}{2}

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

Compute

c :

c = 14^{2} - \frac{7}{2}^{2} : \frac{21}{2}

(- 5, - 3 + \frac{21}{2}), (- 5, - 3 - \frac{21}{2})

What is the foci 4x^2+y^2+40x+6y+95=0 ?
The foci 4x^2+y^2+40x+6y+95=0 is (-5,-3+sqrt(21/2)),(-5,-3-sqrt(21/2))