What is the foci ((x+3)^2)/(21)+((y-6)^2)/(25)=1 ?

The foci ((x+3)^2)/(21)+((y-6)^2)/(25)=1 is (-3,8),(-3,4)

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foci ((x+3)^2)/(21)+((y-6)^2)/(25)=1

foci

\frac{( x + 3 ) ^{2}}{21} + \frac{( y - 6 ) ^{2}}{25} = 1

(- 3, 8), (- 3, 4)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{21} + \frac{( y - 6 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (- 3, 6), b = 5, a = 21

(- 3, 6 + c), (- 3, 6 - c)

Compute

c :

c = 5^{2} - 21^{2} : 2

(- 3, 6 + 2), (- 3, 6 - 2)

Refine

(- 3, 8), (- 3, 4)

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

\frac{( x - 1 ) ^{2}}{9} + \frac{( y + 4 ) ^{2}}{25} = 1

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{81} = 1

16 xx + 25 yy = 400

What is the foci ((x+3)^2)/(21)+((y-6)^2)/(25)=1 ?
The foci ((x+3)^2)/(21)+((y-6)^2)/(25)=1 is (-3,8),(-3,4)