What is the foci 16x^2+25y^2-128x+50y-119=0 ?

The foci 16x^2+25y^2-128x+50y-119=0 is (7,-1),(1,-1)

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foci 16x^2+25y^2-128x+50y-119=0

foci

16 x^{2} + 25 y^{2} - 128 x + 50 y - 119 = 0

(7, - 1), (1, - 1)

Solution steps

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

Calculate ellipse properties

16 x^{2} + 25 y^{2} - 128 x + 50 y - 119 = 0 :

Ellipse with center

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

(4 + c, - 1), (4 - c, - 1)

Compute

c :

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

(4 + 3, - 1), (4 - 3, - 1)

Refine

(7, - 1), (1, - 1)

\frac{x ^{2}}{4} + \frac{y ^{2}}{12} = 1

\frac{( x + 0 ) ^{2}}{100} + \frac{( y + 0 ) ^{2}}{64} = 1

2205 = 180 x^{2} + 196 y^{2}

\frac{x ^{2}}{0.38 1 ^{2}} + \frac{y ^{2}}{0.372 8 ^{2}} = 1

25 x^{2} + 9 y^{2} + 150 x = 0

What is the foci 16x^2+25y^2-128x+50y-119=0 ?
The foci 16x^2+25y^2-128x+50y-119=0 is (7,-1),(1,-1)