What is the foci 1=((x-2)^2)/(16)+((y-3)^2)/(25) ?

The foci 1=((x-2)^2)/(16)+((y-3)^2)/(25) is (2,6),(2,0)

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foci 1=((x-2)^2)/(16)+((y-3)^2)/(25)

foci

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

(2, 6), (2, 0)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (2, 3), b = 5, a = 4

(2, 3 + c), (2, 3 - c)

Compute

c :

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

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

Refine

(2, 6), (2, 0)

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

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

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

16 x^{2} + 9 y^{2} - 144 = 0

x^{2} + 3 y^{2} = 9

What is the foci 1=((x-2)^2)/(16)+((y-3)^2)/(25) ?
The foci 1=((x-2)^2)/(16)+((y-3)^2)/(25) is (2,6),(2,0)