What is the foci (x+6)^2+((y+4)^2)/(1/4)=1 ?

The foci (x+6)^2+((y+4)^2)/(1/4)=1 is ((-12+sqrt(3))/2 ,-4),((-12-sqrt(3))/2 ,-4)

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foci (x+6)^2+((y+4)^2)/(1/4)=1

foci

(\frac{- 12 + 3}{2}, - 4), (\frac{- 12 - 3}{2}, - 4)

Solution steps

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

Calculate ellipse properties

(x + 6)^{2} + \frac{( y + 4 ) ^{2}}{\frac{1}{4}} = 1 :

Ellipse with center

(h, k) = (- 6, - 4), a = 1, b = \frac{1}{2}

(- 6 + c, - 4), (- 6 - c, - 4)

Compute

c :

c = 1^{2} - (\frac{1}{2})^{2} : \frac{3}{2}

(- 6 + \frac{3}{2}, - 4), (- 6 - \frac{3}{2}, - 4)

Refine

(\frac{- 12 + 3}{2}, - 4), (\frac{- 12 - 3}{2}, - 4)

What is the foci (x+6)^2+((y+4)^2)/(1/4)=1 ?
The foci (x+6)^2+((y+4)^2)/(1/4)=1 is ((-12+sqrt(3))/2 ,-4),((-12-sqrt(3))/2 ,-4)