What is the foci ((x-1)^2)/8+((y-3)^2)/4 =1 ?

The foci ((x-1)^2)/8+((y-3)^2)/4 =1 is (3,3),(-1,3)

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foci ((x-1)^2)/8+((y-3)^2)/4 =1

foci

\frac{( x - 1 ) ^{2}}{8} + \frac{( y - 3 ) ^{2}}{4} = 1

(3, 3), (- 1, 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{8} + \frac{( y - 3 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (1, 3), a = 22, b = 2

(1 + c, 3), (1 - c, 3)

Compute

c :

c = (22)^{2} - 2^{2} : 2

(1 + 2, 3), (1 - 2, 3)

Refine

(3, 3), (- 1, 3)

\frac{x ^{2}}{4} + \frac{y ^{2}}{20} = 1

\frac{( x 8 ) ^{2}}{49} + \frac{( y + 1 ) ^{2}}{100} = 1

\frac{x ^{2}}{32} + \frac{y ^{2}}{18} = 1

x^{2} + \frac{y ^{2}}{100} = 1

\frac{( x - 2 ) ^{2}}{8 ^{2}} + \frac{( y + 2 ) ^{2}}{3 ^{2}} = 1

What is the foci ((x-1)^2)/8+((y-3)^2)/4 =1 ?
The foci ((x-1)^2)/8+((y-3)^2)/4 =1 is (3,3),(-1,3)