What is the foci ((x-2)^2)/(64)+((y-4)^2)/(16)=1 ?

The foci ((x-2)^2)/(64)+((y-4)^2)/(16)=1 is (2+4sqrt(3),4),(2-4sqrt(3),4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-2)^2)/(64)+((y-4)^2)/(16)=1

foci

\frac{( x - 2 ) ^{2}}{64} + \frac{( y - 4 ) ^{2}}{16} = 1

(2 + 43, 4), (2 - 43, 4)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{64} + \frac{( y - 4 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (2, 4), a = 8, b = 4

(2 + c, 4), (2 - c, 4)

Compute

c :

c = 8^{2} - 4^{2} : 43

(2 + 43, 4), (2 - 43, 4)

49 x^{2} + 25 y^{2} - 1225 = 0

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

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( X - 1 ) ^{2}}{4} + \frac{( Y + 2 ) ^{2}}{16} = 1

25 x^{2} + 9 y^{2} - 54 y - 144 = 0

What is the foci ((x-2)^2)/(64)+((y-4)^2)/(16)=1 ?
The foci ((x-2)^2)/(64)+((y-4)^2)/(16)=1 is (2+4sqrt(3),4),(2-4sqrt(3),4)