What is the foci ((x-1)^2)/(425)+((y-2)^2)/(400)=1 ?

The foci ((x-1)^2)/(425)+((y-2)^2)/(400)=1 is (6,2),(-4,2)

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foci ((x-1)^2)/(425)+((y-2)^2)/(400)=1

foci

\frac{( x - 1 ) ^{2}}{425} + \frac{( y - 2 ) ^{2}}{400} = 1

(6, 2), (- 4, 2)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{425} + \frac{( y - 2 ) ^{2}}{400} = 1 :

Ellipse with center

(h, k) = (1, 2), a = 517, b = 20

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

Compute

c :

c = (517)^{2} - 2 0^{2} : 5

(1 + 5, 2), (1 - 5, 2)

Refine

(6, 2), (- 4, 2)

9 x^{2} + 4 y^{2} + 18 x - 8 y - 131 = 0

\frac{( x - 2 ) ^{2}}{225} + \frac{( y - 8 ) ^{2}}{81} = 1

x^{2} + 2 y^{2} + 2 x - 20 y + 43 = 0

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

2 x^{2} + 3 y^{2} + 10 x + 10 y - 10 = 0

What is the foci ((x-1)^2)/(425)+((y-2)^2)/(400)=1 ?
The foci ((x-1)^2)/(425)+((y-2)^2)/(400)=1 is (6,2),(-4,2)