What is the foci (x^2)/(121)+((y+4)^2)/(23)=1 ?

The foci (x^2)/(121)+((y+4)^2)/(23)=1 is (7sqrt(2),-4),(-7sqrt(2),-4)

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foci (x^2)/(121)+((y+4)^2)/(23)=1

foci

\frac{x ^{2}}{121} + \frac{( y + 4 ) ^{2}}{23} = 1

(72, - 4), (- 72, - 4)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{121} + \frac{( y + 4 ) ^{2}}{23} = 1 :

Ellipse with center

(h, k) = (0, - 4), a = 11, b = 23

(0 + c, - 4), (0 - c, - 4)

Compute

c :

c = 1 1^{2} - 23^{2} : 72

(0 + 72, - 4), (0 - 72, - 4)

Refine

(72, - 4), (- 72, - 4)

\frac{x ^{2} + y ^{2}}{5} = 1

\frac{( x - 2 ) ^{2}}{16} + \frac{y ^{2}}{9} = 1

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1

16 x^{2} + 25 y^{2} - 96 x + 100 y - 156 = 0

+ \frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

What is the foci (x^2)/(121)+((y+4)^2)/(23)=1 ?
The foci (x^2)/(121)+((y+4)^2)/(23)=1 is (7sqrt(2),-4),(-7sqrt(2),-4)