What is the foci (x^2}{49}+\frac{y^2)/9 =1 ?

The foci (x^2}{49}+\frac{y^2)/9 =1 is (2sqrt(10),0),(-2sqrt(10),0)

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foci (x^2}{49}+\frac{y^2)/9 =1

foci

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

(210, 0), (- 210, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = 7, b = 3

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

Compute

c :

c = 7^{2} - 3^{2} : 210

(0 + 210, 0), (0 - 210, 0)

Refine

(210, 0), (- 210, 0)

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

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

\frac{x ^{2}}{5} + \frac{y ^{2}}{3} = 1

6 x^{2} + 8 y^{2} + 12 x - 32 y = 178

9 x^{2} + 49 y^{2} - 144 x + 490 y + 1360 = 0

What is the foci (x^2}{49}+\frac{y^2)/9 =1 ?
The foci (x^2}{49}+\frac{y^2)/9 =1 is (2sqrt(10),0),(-2sqrt(10),0)