What is the foci ((x+1)^2)/(64)+((y-1)^2)/(100)=1 ?

The foci ((x+1)^2)/(64)+((y-1)^2)/(100)=1 is (-1,7),(-1,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+1)^2)/(64)+((y-1)^2)/(100)=1

foci

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

(- 1, 7), (- 1, - 5)

Solution steps

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

Calculate ellipse properties

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

Ellipse with center

(h, k) = (- 1, 1), b = 10, a = 8

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

Compute

c :

c = 1 0^{2} - 8^{2} : 6

(- 1, 1 + 6), (- 1, 1 - 6)

Refine

(- 1, 7), (- 1, - 5)

x^{2} + 16 y^{2} - 64 = 0

18 x^{2} + 24 x + 12 y^{2} - 12 y = 1

x^{2} + 25 y^{2} = 25

x^{2} + 9 y^{2} + 6 x - 90 y + 225 = 0

x^{2} + 3 y^{2} - 4 x + 12 y + 13 = 0

What is the foci ((x+1)^2)/(64)+((y-1)^2)/(100)=1 ?
The foci ((x+1)^2)/(64)+((y-1)^2)/(100)=1 is (-1,7),(-1,-5)