What is the foci ((x-1)^2)/(36)+((y+2)^2)/(25)=1 ?

The foci ((x-1)^2)/(36)+((y+2)^2)/(25)=1 is (1+sqrt(11),-2),(1-sqrt(11),-2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-1)^2)/(36)+((y+2)^2)/(25)=1

foci

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

(1 + 11, - 2), (1 - 11, - 2)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (1, - 2), a = 6, b = 5

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

Compute

c :

c = 6^{2} - 5^{2} : 11

(1 + 11, - 2), (1 - 11, - 2)

36 x^{2} + 9 y^{2} + 36 y - 288 = 0

100 x^{2} + 36 y^{2} = 225

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

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

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

What is the foci ((x-1)^2)/(36)+((y+2)^2)/(25)=1 ?
The foci ((x-1)^2)/(36)+((y+2)^2)/(25)=1 is (1+sqrt(11),-2),(1-sqrt(11),-2)