What is the foci (x^2)/7+(y^2)/(43)=1 ?

The foci (x^2)/7+(y^2)/(43)=1 is (0,6),(0,-6)

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foci (x^2)/7+(y^2)/(43)=1

foci

\frac{x ^{2}}{7} + \frac{y ^{2}}{43} = 1

(0, 6), (0, - 6)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{7} + \frac{y ^{2}}{43} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 43, a = 7

(0, 0 + c), (0, 0 - c)

Compute

c :

c = 43^{2} - 7^{2} : 6

(0, 0 + 6), (0, 0 - 6)

Refine

(0, 6), (0, - 6)

3 x^{2} + 10 y^{2} = 30

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

\frac{x ^{2}}{81} + \frac{y ^{2}}{225} = 1

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

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 4 ) ^{2}}{36} = 1

What is the foci (x^2)/7+(y^2)/(43)=1 ?
The foci (x^2)/7+(y^2)/(43)=1 is (0,6),(0,-6)