What is the foci 25x^2+9y^2+50x-108y+124=0 ?

The foci 25x^2+9y^2+50x-108y+124=0 is (-1,10),(-1,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci 25x^2+9y^2+50x-108y+124=0

foci

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0

(- 1, 10), (- 1, 2)

Solution steps

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

Calculate ellipse properties

25 x^{2} + 9 y^{2} + 50 x - 108 y + 124 = 0 :

Ellipse with center

(h, k) = (- 1, 6), b = 5, a = 3

(- 1, 6 + c), (- 1, 6 - c)

Compute

c :

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

(- 1, 6 + 4), (- 1, 6 - 4)

Refine

(- 1, 10), (- 1, 2)

x^{2} + 2 y^{2} = 4

\frac{x ^{2}}{3} + \frac{y ^{2}}{5} = 1

16 x^{2} + 25 y^{2} - 64 x + 150 y - 111 = 0

9 x^{2} + 49 y^{2} = 441

\frac{( x - 1 ) ^{2}}{425} + \frac{( y - 2 ) ^{2}}{400} = 1

What is the foci 25x^2+9y^2+50x-108y+124=0 ?
The foci 25x^2+9y^2+50x-108y+124=0 is (-1,10),(-1,2)