What is the foci (x^2)/(49)+(y^2)/(36)=1 ?

The foci (x^2)/(49)+(y^2)/(36)=1 is (sqrt(13),0),(-sqrt(13),0)

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foci (x^2)/(49)+(y^2)/(36)=1

foci

\frac{x ^{2}}{49} + \frac{y ^{2}}{36} = 1

(13, 0), (- 13, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{49} + \frac{y ^{2}}{36} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 7, b = 6

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

Compute

c :

c = 7^{2} - 6^{2} : 13

(0 + 13, 0), (0 - 13, 0)

Refine

(13, 0), (- 13, 0)

x = - \frac{1}{8} y^{2}

x^{2} + y^{2} = 13

\frac{x ^{2}}{64} + \frac{y ^{2}}{36} = 1

- 1 = x^{2} - y^{2}

y^{2} = - 12 x

What is the foci (x^2)/(49)+(y^2)/(36)=1 ?
The foci (x^2)/(49)+(y^2)/(36)=1 is (sqrt(13),0),(-sqrt(13),0)