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

The foci 2x^2+4y^2-8x-4y-15=0 is (2+sqrt(6), 1/2),(2-sqrt(6), 1/2)

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

foci

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

(2 + 6, \frac{1}{2}), (2 - 6, \frac{1}{2})

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, \frac{1}{2}), a = 23, b = 6

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

Compute

c :

c = (23)^{2} - 6^{2} : 6

(2 + 6, \frac{1}{2}), (2 - 6, \frac{1}{2})

3 x^{2} + y^{2} = 9

16 x^{2} + 81 y^{2} = 81

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

\frac{( x + 3 ) ^{2}}{49} + \frac{y ^{2}}{36} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

What is the foci 2x^2+4y^2-8x-4y-15=0 ?
The foci 2x^2+4y^2-8x-4y-15=0 is (2+sqrt(6), 1/2),(2-sqrt(6), 1/2)