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

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

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foci (x^2}{49}+\frac{y^2)/4 =1

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

\frac{x ^{2}}{3. 2 ^{2}} + \frac{y ^{2}}{2.4 8 ^{2}} = 1

\frac{( x - 3 ) ^{2}}{16} + \frac{( y + 1 ) ^{2}}{9} = 1

5 x^{2} + 2 y^{2} = 10

\frac{x ^{2}}{36} + \frac{y ^{2}}{81} = 1

\frac{y ^{2}}{81} + \frac{x ^{2}}{36} = 1

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