What is the foci ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1 ?

The foci ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1 is (9,-3),(1,-3)

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foci ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1

foci

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{9} = 1

(9, - 3), (1, - 3)

Solution steps

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

Calculate ellipse properties

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (5, - 3), a = 5, b = 3

(5 + c, - 3), (5 - c, - 3)

Compute

c :

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

(5 + 4, - 3), (5 - 4, - 3)

Refine

(9, - 3), (1, - 3)

x^{2} + y^{2} = 7

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

x^{2} = 12 y

\frac{x ^{2}}{16} + \frac{y ^{2}}{36} = 1

f (x) = 2 x^{2} + 28 x + 105

What is the foci ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1 ?
The foci ((x-5)^2}{25}+\frac{(y+3)^2)/9 =1 is (9,-3),(1,-3)