What is the foci ((x-1)^2)/(45)+((y-2)^2)/(20)=1 ?

The foci ((x-1)^2)/(45)+((y-2)^2)/(20)=1 is (6,2),(-4,2)

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foci ((x-1)^2)/(45)+((y-2)^2)/(20)=1

foci

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 2 ) ^{2}}{20} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 2 ) ^{2}}{20} = 1 :

Ellipse with center

(h, k) = (1, 2), a = 35, b = 25

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

Compute

c :

c = (35)^{2} - (25)^{2} : 5

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

Refine

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

\frac{x ^{2}}{100} + \frac{y ^{2}}{16} = 1

\frac{( x + 4 ) ^{2}}{5} + \frac{( y - 5 ) ^{2}}{9} = 1

\frac{x ^{2}}{1} + \frac{y ^{2}}{64} = 1

\frac{( x ) ^{2}}{13} + \frac{y ^{2}}{4} = 1

\frac{( x - 3 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{16} = 1

What is the foci ((x-1)^2)/(45)+((y-2)^2)/(20)=1 ?
The foci ((x-1)^2)/(45)+((y-2)^2)/(20)=1 is (6,2),(-4,2)