What is the foci 16(x-4)^2+25(y+2)^2=400 ?

The foci 16(x-4)^2+25(y+2)^2=400 is (7,-2),(1,-2)

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foci 16(x-4)^2+25(y+2)^2=400

foci

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400

(7, - 2), (1, - 2)

Solution steps

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

Calculate ellipse properties

16 (x - 4)^{2} + 25 (y + 2)^{2} = 400 :

Ellipse with center

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

(4 + c, - 2), (4 - c, - 2)

Compute

c :

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

(4 + 3, - 2), (4 - 3, - 2)

Refine

(7, - 2), (1, - 2)

\frac{x ^{2}}{38} + \frac{y ^{2}}{22} = 1

\frac{( x - 4 ) ^{2}}{25} + \frac{( y - 2 ) ^{2}}{16} = 1

14 x^{2} + 5 y^{2} - 70 = 0

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

16 x^{2} + 25 y^{2} + 100 y = 300

What is the foci 16(x-4)^2+25(y+2)^2=400 ?
The foci 16(x-4)^2+25(y+2)^2=400 is (7,-2),(1,-2)