What is the foci 16x^2+25y^2-64x-50y-311=0 ?

The foci 16x^2+25y^2-64x-50y-311=0 is (5,1),(-1,1)

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foci 16x^2+25y^2-64x-50y-311=0

foci

16 x^{2} + 25 y^{2} - 64 x - 50 y - 311 = 0

(5, 1), (- 1, 1)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} - 64 x - 50 y - 311 = 0 :

Ellipse with center

(h, k) = (2, 1), a = 5, b = 4

(2 + c, 1), (2 - c, 1)

Compute

c :

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

(2 + 3, 1), (2 - 3, 1)

Refine

(5, 1), (- 1, 1)

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 2 ) ^{2}}{4} = 1

\frac{x ^{2}}{3} + \frac{y ^{2}}{2} = 1

\frac{x ^{2}}{16} + \frac{y ^{2}}{81} = 1

\frac{( x - 8 ) ^{2}}{49} + \frac{( y + 5 ) ^{2}}{9} = 1

\frac{( x - 4 ) ^{2}}{4} + \frac{( y - 3 ) ^{2}}{9} = 1

What is the foci 16x^2+25y^2-64x-50y-311=0 ?
The foci 16x^2+25y^2-64x-50y-311=0 is (5,1),(-1,1)