What is the foci (x^2)/(38)+(y^2)/(22)=1 ?

The foci (x^2)/(38)+(y^2)/(22)=1 is (4,0),(-4,0)

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foci (x^2)/(38)+(y^2)/(22)=1

foci

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

(4, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = 38, b = 22

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

Compute

c :

c = 38^{2} - 22^{2} : 4

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

\frac{( x - 4 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{16} = 1

14 x^{2} + 5 y^{2} - 70 = 0

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

16 x^{2} + 25 y^{2} + 100 y = 300

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 3 ) ^{2}}{9} = 1

What is the foci (x^2)/(38)+(y^2)/(22)=1 ?
The foci (x^2)/(38)+(y^2)/(22)=1 is (4,0),(-4,0)