What is the foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1 ?

The foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1 is (-3,8),(-3,2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1

foci

(- 3, 8), (- 3, 2)

Solution steps

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

Calculate ellipse properties

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1 :

Ellipse with center

(h, k) = (- 3, 5), b = 5, a = 4

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

Compute

c :

c = 5^{2} - 4^{2} : 3

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

Refine

(- 3, 8), (- 3, 2)

What is the foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1 ?
The foci ((x+3)^2)/(16)+((y-5)^2)/(25)=1 is (-3,8),(-3,2)