What is the foci (x^2)/(67)+(y^2)/(57)=1 ?

The foci (x^2)/(67)+(y^2)/(57)=1 is (sqrt(10),0),(-sqrt(10),0)

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foci (x^2)/(67)+(y^2)/(57)=1

foci

\frac{x ^{2}}{67} + \frac{y ^{2}}{57} = 1

(10, 0), (- 10, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{67} + \frac{y ^{2}}{57} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 67, b = 57

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

Compute

c :

c = 67^{2} - 57^{2} : 10

(0 + 10, 0), (0 - 10, 0)

Refine

(10, 0), (- 10, 0)

4 x^{2} + 16 y^{2} = 64

\frac{( x + 4 ) ^{2}}{22} + \frac{( y + 3 ) ^{2}}{26} = 1

+ 4 x^{2} + 9 y^{2} - 16 x + 18 y = 11

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

\frac{x ^{2}}{38} + \frac{y ^{2}}{22} = 1

What is the foci (x^2)/(67)+(y^2)/(57)=1 ?
The foci (x^2)/(67)+(y^2)/(57)=1 is (sqrt(10),0),(-sqrt(10),0)