What is the foci (x^2)/(17)+(y^2)/(35)=1 ?

The foci (x^2)/(17)+(y^2)/(35)=1 is (0,3sqrt(2)),(0,-3sqrt(2))

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foci (x^2)/(17)+(y^2)/(35)=1

foci

\frac{x ^{2}}{17} + \frac{y ^{2}}{35} = 1

(0, 32), (0, - 32)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{17} + \frac{y ^{2}}{35} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 35, a = 17

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

Compute

c :

c = 35^{2} - 17^{2} : 32

(0, 0 + 32), (0, 0 - 32)

Refine

(0, 32), (0, - 32)

64 x^{2} + 81 y^{2} = 5184

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

4 x^{2} + y^{2} + 40 x + 6 y + 95 = 0

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

\frac{( y + 4 ) ^{2}}{24} + \frac{( x - 3 ) ^{2}}{12} = 1

What is the foci (x^2)/(17)+(y^2)/(35)=1 ?
The foci (x^2)/(17)+(y^2)/(35)=1 is (0,3sqrt(2)),(0,-3sqrt(2))