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

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

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foci 16(x-1)^2+25(y-4)^2=400

foci

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

(4, 4), (- 2, 4)

Solution steps

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

Calculate ellipse properties

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400 :

Ellipse with center

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

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

Compute

c :

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

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

Refine

(4, 4), (- 2, 4)

\frac{x ^{2}}{121} + \frac{( y + 4 ) ^{2}}{1} = 1

9 x^{2} + 5 y^{2} = 45

16 x^{2} + 25 y^{2} + 64 x - 336 = 0

\frac{( x - 7 ) ^{2}}{81} + \frac{( y - 6 ) ^{2}}{64} = 1

\frac{( X + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

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