What is the foci 9(x-5)^2+4(y+2)^2=36 ?

The foci 9(x-5)^2+4(y+2)^2=36 is (5,-2+sqrt(5)),(5,-2-sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci 9(x-5)^2+4(y+2)^2=36

foci

9 (x - 5)^{2} + 4 (y + 2)^{2} = 36

(5, - 2 + 5), (5, - 2 - 5)

Solution steps

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

Calculate ellipse properties

9 (x - 5)^{2} + 4 (y + 2)^{2} = 36 :

Ellipse with center

(h, k) = (5, - 2), b = 3, a = 2

(5, - 2 + c), (5, - 2 - c)

Compute

c :

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

(5, - 2 + 5), (5, - 2 - 5)

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

\frac{x ^{2}}{9} + \frac{y ^{2}}{3} = 1

y^{2} = 1 - 25 x^{2}

2 x^{2} - 8 x + 3 y^{2} + 6 y + 5 = 0

36 x^{2} + 288 x + 64 y^{2} - 256 y = 1472

What is the foci 9(x-5)^2+4(y+2)^2=36 ?
The foci 9(x-5)^2+4(y+2)^2=36 is (5,-2+sqrt(5)),(5,-2-sqrt(5))