What is the foci (x^2}{121}+\frac{(y+4)^2)/1 =1 ?

The foci (x^2}{121}+\frac{(y+4)^2)/1 =1 is (2sqrt(30),-4),(-2sqrt(30),-4)

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

foci

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

(230, - 4), (- 230, - 4)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (0, - 4), a = 11, b = 1

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

Compute

c :

c = 1 1^{2} - 1^{2} : 230

(0 + 230, - 4), (0 - 230, - 4)

Refine

(230, - 4), (- 230, - 4)

9 x^{2} + 5 y^{2} = 45

16 x^{2} + 25 y^{2} + 64 x - 336 = 0

\frac{( x - 7 ) ^{2}}{81} + \frac{( y - 6 ) ^{2}}{64} = 1

\frac{( X + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

6 x^{2} + 10 y^{2} = 60

What is the foci (x^2}{121}+\frac{(y+4)^2)/1 =1 ?
The foci (x^2}{121}+\frac{(y+4)^2)/1 =1 is (2sqrt(30),-4),(-2sqrt(30),-4)