What is the foci 16x^2+25y^2-48x+100y-264=0 ?

The foci 16x^2+25y^2-48x+100y-264=0 is (9/2 ,-2),(-3/2 ,-2)

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foci 16x^2+25y^2-48x+100y-264=0

foci

16 x^{2} + 25 y^{2} - 48 x + 100 y - 264 = 0

(\frac{9}{2}, - 2), (- \frac{3}{2}, - 2)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} - 48 x + 100 y - 264 = 0 :

Ellipse with center

(h, k) = (\frac{3}{2}, - 2), a = 5, b = 4

(\frac{3}{2} + c, - 2), (\frac{3}{2} - c, - 2)

Compute

c :

c = 5^{2} - 4^{2} : 3

(\frac{3}{2} + 3, - 2), (\frac{3}{2} - 3, - 2)

Refine

(\frac{9}{2}, - 2), (- \frac{3}{2}, - 2)

9 x^{2} + 25 y^{2} + 54 x + 100 y - 44 = 0

15 x^{2} + 16 y^{2} = 240

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 2 ) ^{2}}{20} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{16} = 1

What is the foci 16x^2+25y^2-48x+100y-264=0 ?
The foci 16x^2+25y^2-48x+100y-264=0 is (9/2 ,-2),(-3/2 ,-2)