What is the foci ((x+2)^2)/9+((y-1)^2)/(16)=1 ?

The foci ((x+2)^2)/9+((y-1)^2)/(16)=1 is (-2,1+sqrt(7)),(-2,1-sqrt(7))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+2)^2)/9+((y-1)^2)/(16)=1

foci

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{16} = 1

(- 2, 1 + 7), (- 2, 1 - 7)

Solution steps

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

Calculate ellipse properties

\frac{( x + 2 ) ^{2}}{9} + \frac{( y - 1 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (- 2, 1), b = 4, a = 3

(- 2, 1 + c), (- 2, 1 - c)

Compute

c :

c = 4^{2} - 3^{2} : 7

(- 2, 1 + 7), (- 2, 1 - 7)

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

25 x^{2} + 144 y^{2} = 900

3 x^{2} + 6 y^{2} = 18

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

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

What is the foci ((x+2)^2)/9+((y-1)^2)/(16)=1 ?
The foci ((x+2)^2)/9+((y-1)^2)/(16)=1 is (-2,1+sqrt(7)),(-2,1-sqrt(7))