What is the foci (x^2)/(32)+(y^2)/(16)=1 ?

The foci (x^2)/(32)+(y^2)/(16)=1 is (4,0),(-4,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci (x^2)/(32)+(y^2)/(16)=1

foci

\frac{x ^{2}}{32} + \frac{y ^{2}}{16} = 1

(4, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{32} + \frac{y ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 42, b = 4

(0 + c, 0), (0 - c, 0)

Compute

c :

c = (42)^{2} - 4^{2} : 4

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

\frac{x ^{2}}{100} + \frac{( y - 6 ) ^{2}}{64} = 1

6 x^{2} + 9 y^{2} - 18 y - 45 = 0

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

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

4 x^{2} + 9 y^{2} + 16 x - 90 y + 205 = 0

What is the foci (x^2)/(32)+(y^2)/(16)=1 ?
The foci (x^2)/(32)+(y^2)/(16)=1 is (4,0),(-4,0)