What is the foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1 ?

The foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1 is (-3+3sqrt(3),1),(-3-3sqrt(3),1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1

foci

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{9} = 1

(- 3 + 33, 1), (- 3 - 33, 1)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 1 ) ^{2}}{9} = 1 :

Ellipse with center

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

(- 3 + c, 1), (- 3 - c, 1)

Compute

c :

c = 6^{2} - 3^{2} : 33

(- 3 + 33, 1), (- 3 - 33, 1)

2 x^{2} + 3 y^{2} + 4 x = 1

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

\frac{( x - 1 ) ^{2}}{25} + \frac{y ^{2}}{16} = 1

\frac{x ^{2}}{576} + \frac{y ^{2}}{676} = 1

x^{2} + 3 y^{2} + 6 x + 6 = 0

What is the foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1 ?
The foci ((x+3)^2}{36}+\frac{(y-1)^2)/9 =1 is (-3+3sqrt(3),1),(-3-3sqrt(3),1)