What is the foci 9x^2+49y^2=441 ?

The foci 9x^2+49y^2=441 is (2sqrt(10),0),(-2sqrt(10),0)

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foci 9x^2+49y^2=441

foci

9 x^{2} + 49 y^{2} = 441

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

Solution steps

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

Calculate ellipse properties

9 x^{2} + 49 y^{2} = 441 :

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 - 1 ) ^{2}}{425} + \frac{( y - 2 ) ^{2}}{400} = 1

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0

\frac{( x - 2 ) ^{2}}{225} + \frac{( y - 8 ) ^{2}}{81} = 1

x^{2} + 2 y^{2} + 2 x - 20 y + 43 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

What is the foci 9x^2+49y^2=441 ?
The foci 9x^2+49y^2=441 is (2sqrt(10),0),(-2sqrt(10),0)