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

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

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foci (x^2)/k+(y^2)/(4+k)=1

foci

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

(2 i, 0), (- 2 i, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, 0), a = k, b = 4 + k

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

Compute

c :

c = k^{2} - 4 + k^{2} : 2 i

(0 + 2 i, 0), (0 - 2 i, 0)

Refine

(2 i, 0), (- 2 i, 0)

16 x^{2} + 25 y^{2} - 128 x + 50 y - 119 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{12} = 1

\frac{( x + 0 ) ^{2}}{100} + \frac{( y + 0 ) ^{2}}{64} = 1

2205 = 180 x^{2} + 196 y^{2}

\frac{x ^{2}}{0.38 1 ^{2}} + \frac{y ^{2}}{0.372 8 ^{2}} = 1

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