What is the foci ((x-1)^2)/6+((y+2)^2)/(18)=1 ?

The foci ((x-1)^2)/6+((y+2)^2)/(18)=1 is (1,-2+2sqrt(3)),(1,-2-2sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-1)^2)/6+((y+2)^2)/(18)=1

foci

\frac{( x - 1 ) ^{2}}{6} + \frac{( y + 2 ) ^{2}}{18} = 1

(1, - 2 + 23), (1, - 2 - 23)

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{6} + \frac{( y + 2 ) ^{2}}{18} = 1 :

Ellipse with center

(h, k) = (1, - 2), b = 32, a = 6

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

Compute

c :

c = (32)^{2} - 6^{2} : 23

(1, - 2 + 23), (1, - 2 - 23)

\frac{( x + 11 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{1 9 ^{2}} = 1

\frac{( x + 6 ) ^{2}}{18} + \frac{( y + 3 ) ^{2}}{9} = 1

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

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

16 x^{2} + 25 y^{2} - 128 x + 50 y - 119 = 0

What is the foci ((x-1)^2)/6+((y+2)^2)/(18)=1 ?
The foci ((x-1)^2)/6+((y+2)^2)/(18)=1 is (1,-2+2sqrt(3)),(1,-2-2sqrt(3))