What is the foci x^2+8(y+5)^2=20 ?

The foci x^2+8(y+5)^2=20 is (sqrt(35/2),-5),(-sqrt(35/2),-5)

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foci x^2+8(y+5)^2=20

foci

x^{2} + 8 (y + 5)^{2} = 20

(\frac{35}{2}, - 5), (- \frac{35}{2}, - 5)

Solution steps

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

Calculate ellipse properties

x^{2} + 8 (y + 5)^{2} = 20 :

Ellipse with center

(h, k) = (0, - 5), a = 25, b = \frac{5}{2}

(0 + c, - 5), (0 - c, - 5)

Compute

c :

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

(0 + \frac{35}{2}, - 5), (0 - \frac{35}{2}, - 5)

Refine

(\frac{35}{2}, - 5), (- \frac{35}{2}, - 5)

\frac{x ^{2}}{9} + \frac{y ^{2}}{16} > 1

4 x^{2} + 9 y^{2} + 8 x - 36 y + 4 = 0

\frac{x ^{2}}{25} + \frac{y ^{2}}{36} = 1

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

9 x^{2} + 16 y^{2} - 54 x + 64 y + 1 = 0

What is the foci x^2+8(y+5)^2=20 ?
The foci x^2+8(y+5)^2=20 is (sqrt(35/2),-5),(-sqrt(35/2),-5)