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

The foci 4x^2+y^2-48x+4y+48=0 is (6,-2+5sqrt(3)),(6,-2-5sqrt(3))

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

foci

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

(6, - 2 + 53), (6, - 2 - 53)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (6, - 2), b = 10, a = 5

(6, - 2 + c), (6, - 2 - c)

Compute

c :

c = 1 0^{2} - 5^{2} : 53

(6, - 2 + 53), (6, - 2 - 53)

7 x^{2} = 14 - 2 y^{2}

25 x^{2} + 4 y^{2} - 100 = 0

\frac{x ^{2}}{1.586 9 ^{2}} + \frac{y ^{2}}{0.9989} = 1

\frac{x ^{2}}{36} + \frac{y ^{2}}{49} = 1

\frac{( x + 7 ) ^{2}}{90} + \frac{( y + 8 ) ^{2}}{84} = 1

What is the foci 4x^2+y^2-48x+4y+48=0 ?
The foci 4x^2+y^2-48x+4y+48=0 is (6,-2+5sqrt(3)),(6,-2-5sqrt(3))