What is the foci 8x^2+4y^2+16x-12y-15=0 ?

The foci 8x^2+4y^2+16x-12y-15=0 is (-1, 7/2),(-1,-1/2)

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foci 8x^2+4y^2+16x-12y-15=0

foci

(- 1, \frac{7}{2}), (- 1, - \frac{1}{2})

Solution steps

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

Calculate ellipse properties

8 x^{2} + 4 y^{2} + 16 x - 12 y - 15 = 0 :

Ellipse with center

(h, k) = (- 1, \frac{3}{2}), b = 22, a = 2

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

Compute

c :

c = (22)^{2} - 2^{2} : 2

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

Refine

(- 1, \frac{7}{2}), (- 1, - \frac{1}{2})

What is the foci 8x^2+4y^2+16x-12y-15=0 ?
The foci 8x^2+4y^2+16x-12y-15=0 is (-1, 7/2),(-1,-1/2)