What is the foci ((x+7)^2)/8+((y-10)^2)/(13)=1 ?

The foci ((x+7)^2)/8+((y-10)^2)/(13)=1 is (-7,10+sqrt(5)),(-7,10-sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+7)^2)/8+((y-10)^2)/(13)=1

foci

\frac{( x + 7 ) ^{2}}{8} + \frac{( y - 10 ) ^{2}}{13} = 1

(- 7, 10 + 5), (- 7, 10 - 5)

Solution steps

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

Calculate ellipse properties

\frac{( x + 7 ) ^{2}}{8} + \frac{( y - 10 ) ^{2}}{13} = 1 :

Ellipse with center

(h, k) = (- 7, 10), b = 13, a = 22

(- 7, 10 + c), (- 7, 10 - c)

Compute

c :

c = 13^{2} - (22)^{2} : 5

(- 7, 10 + 5), (- 7, 10 - 5)

\frac{( x - 1 ) ^{2}}{49} + \frac{( y - 3 ) ^{2}}{36} = 1

4 x^{2} + y^{2} - 32 x + 16 y + 124 = 0

1 x^{2} + 3 y^{2} + 10 x + 8 y - 10 = 0

4 x^{2} + y^{2} - 8 x + 4 y + 4 = 0

\frac{x ^{2}}{16} + \frac{y ^{2}}{64} = 1

What is the foci ((x+7)^2)/8+((y-10)^2)/(13)=1 ?
The foci ((x+7)^2)/8+((y-10)^2)/(13)=1 is (-7,10+sqrt(5)),(-7,10-sqrt(5))