What is the foci (x^2)/4+(y^2)/(49)=1 ?

The foci (x^2)/4+(y^2)/(49)=1 is (0,3sqrt(5)),(0,-3sqrt(5))

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foci (x^2)/4+(y^2)/(49)=1

foci

\frac{x ^{2}}{4} + \frac{y ^{2}}{49} = 1

(0, 35), (0, - 35)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{y ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (0, 0), b = 7, a = 2

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

Compute

c :

c = 7^{2} - 2^{2} : 35

(0, 0 + 35), (0, 0 - 35)

Refine

(0, 35), (0, - 35)

\frac{( x + 5 ) ^{2}}{9} + (y - 1)^{2} = 1

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

\frac{x ^{2}}{49} + \frac{y ^{2}}{9} = 1

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 5 ) ^{2}}{9} = 1

7 x^{2} + 16 y^{2} = 112

What is the foci (x^2)/4+(y^2)/(49)=1 ?
The foci (x^2)/4+(y^2)/(49)=1 is (0,3sqrt(5)),(0,-3sqrt(5))