What is the foci (2(x-6)^2)/3+(7(y+4)^2)/4 =1 ?

The foci (2(x-6)^2)/3+(7(y+4)^2)/4 =1 is (6+sqrt(13/14),-4),(6-sqrt(13/14),-4)

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foci (2(x-6)^2)/3+(7(y+4)^2)/4 =1

foci

(6 + \frac{13}{14}, - 4), (6 - \frac{13}{14}, - 4)

Solution steps

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

Calculate ellipse properties

\frac{2 ( x - 6 ) ^{2}}{3} + \frac{7 ( y + 4 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (6, - 4), a = \frac{3}{2}, b = \frac{2}{7}

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

Compute

c :

c = \frac{3}{2}^{2} - (\frac{2}{7})^{2} : \frac{13}{14}

(6 + \frac{13}{14}, - 4), (6 - \frac{13}{14}, - 4)

What is the foci (2(x-6)^2)/3+(7(y+4)^2)/4 =1 ?
The foci (2(x-6)^2)/3+(7(y+4)^2)/4 =1 is (6+sqrt(13/14),-4),(6-sqrt(13/14),-4)