What is the foci ((x+3)^2)/9+((y-5)^2)/3 =1 ?

The foci ((x+3)^2)/9+((y-5)^2)/3 =1 is (-3+sqrt(6),5),(-3-sqrt(6),5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+3)^2)/9+((y-5)^2)/3 =1

foci

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1

(- 3 + 6, 5), (- 3 - 6, 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 5 ) ^{2}}{3} = 1 :

Ellipse with center

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

(- 3 + c, 5), (- 3 - c, 5)

Compute

c :

c = 3^{2} - 3^{2} : 6

(- 3 + 6, 5), (- 3 - 6, 5)

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

16 x^{2} + 25 y^{2} - 128 x + 150 y + 81 = 0

x^{2} + 4 y^{2} + 8 y = 0

16 x^{2} + 25 y^{2} - 96 x - 256 = 0

\frac{( ( x + 7 ) ^{2} )}{9} + \frac{( ( y - 4 ) ^{2} )}{7} = 1

What is the foci ((x+3)^2)/9+((y-5)^2)/3 =1 ?
The foci ((x+3)^2)/9+((y-5)^2)/3 =1 is (-3+sqrt(6),5),(-3-sqrt(6),5)