What is the foci 4x^2+16y^2-16x-48=0 ?

The foci 4x^2+16y^2-16x-48=0 is (2+2sqrt(3),0),(2-2sqrt(3),0)

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foci 4x^2+16y^2-16x-48=0

foci

4 x^{2} + 16 y^{2} - 16 x - 48 = 0

(2 + 23, 0), (2 - 23, 0)

Solution steps

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

Calculate ellipse properties

4 x^{2} + 16 y^{2} - 16 x - 48 = 0 :

Ellipse with center

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

(2 + c, 0), (2 - c, 0)

Compute

c :

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

(2 + 23, 0), (2 - 23, 0)

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

4 x^{2} + 9 y^{2} - 48 x + 72 y = - 144

36 x^{2} + 16 y^{2} - 180 x + 96 y - 207 = 0

x^{2} + 2 y^{2} + 2 x - 4 y - 1 = 0

2 x^{2} + 3 y^{2} = 6

What is the foci 4x^2+16y^2-16x-48=0 ?
The foci 4x^2+16y^2-16x-48=0 is (2+2sqrt(3),0),(2-2sqrt(3),0)