What is the foci (x^2}{140}+\frac{(y-6)^2)/4 =1 ?

The foci (x^2}{140}+\frac{(y-6)^2)/4 =1 is (2sqrt(34),6),(-2sqrt(34),6)

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foci (x^2}{140}+\frac{(y-6)^2)/4 =1

foci

\frac{x ^{2}}{140} + \frac{( y - 6 ) ^{2}}{4} = 1

(234, 6), (- 234, 6)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{140} + \frac{( y - 6 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 6), a = 235, b = 2

(0 + c, 6), (0 - c, 6)

Compute

c :

c = (235)^{2} - 2^{2} : 234

(0 + 234, 6), (0 - 234, 6)

Refine

(234, 6), (- 234, 6)

\frac{x ^{2}}{81} + \frac{y ^{2}}{4} = 1

25 x^{2} + 16 y^{2} - 150 x - 32 y - 159 = 0

\frac{( x + 6 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{25} = 1

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

\frac{x ^{2}}{32} + \frac{y ^{2}}{16} = 1

What is the foci (x^2}{140}+\frac{(y-6)^2)/4 =1 ?
The foci (x^2}{140}+\frac{(y-6)^2)/4 =1 is (2sqrt(34),6),(-2sqrt(34),6)