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

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

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foci (x^2)/9+(y^2)/(49)=1

foci

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

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

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

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

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

Compute

c :

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

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

Refine

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

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

\frac{x ^{2}}{5 ^{2}} + \frac{y ^{2}}{4 ^{2}} = 1

\frac{( x - 1 ) ^{2}}{81} + \frac{( y - 7 ) ^{2}}{45} = 1

3 x^{2} + 7 y^{2} = 21

x^{2} + y^{2} + 10 x - 6 y + 33 = 0

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