What is the foci ((x-4)^2)/4+(y^2)/(16)=1 ?

The foci ((x-4)^2)/4+(y^2)/(16)=1 is (4,2sqrt(3)),(4,-2sqrt(3))

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foci ((x-4)^2)/4+(y^2)/(16)=1

foci

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

(4, 23), (4, - 23)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (4, 0), b = 4, a = 2

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

Compute

c :

c = 4^{2} - 2^{2} : 23

(4, 0 + 23), (4, 0 - 23)

Refine

(4, 23), (4, - 23)

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

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

25 x^{2} + 4 y^{2} = 100

9 x^{2} + 36 y^{2} + 36 x + 432 y = - 1008

4 x^{2} + 9 y^{2} - 32 x - 36 y + 64 = 0

What is the foci ((x-4)^2)/4+(y^2)/(16)=1 ?
The foci ((x-4)^2)/4+(y^2)/(16)=1 is (4,2sqrt(3)),(4,-2sqrt(3))