What is the foci ((x+4)^2)/(24)+((y+2)^2)/(49)=1 ?

The foci ((x+4)^2)/(24)+((y+2)^2)/(49)=1 is (-4,3),(-4,-7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+4)^2)/(24)+((y+2)^2)/(49)=1

foci

\frac{( x + 4 ) ^{2}}{24} + \frac{( y + 2 ) ^{2}}{49} = 1

(- 4, 3), (- 4, - 7)

Solution steps

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

Calculate ellipse properties

\frac{( x + 4 ) ^{2}}{24} + \frac{( y + 2 ) ^{2}}{49} = 1 :

Ellipse with center

(h, k) = (- 4, - 2), b = 7, a = 26

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

Compute

c :

c = 7^{2} - (26)^{2} : 5

(- 4, - 2 + 5), (- 4, - 2 - 5)

Refine

(- 4, 3), (- 4, - 7)

3 x^{2} + 4 y^{2} = 48

9 (x + 2)^{2} + 4 (y - 3)^{2} = 36

\frac{( x + 1 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{169} = 1

16 (x - 1)^{2} + 25 (y - 4)^{2} = 400

\frac{x ^{2}}{121} + \frac{( y + 4 ) ^{2}}{1} = 1

What is the foci ((x+4)^2)/(24)+((y+2)^2)/(49)=1 ?
The foci ((x+4)^2)/(24)+((y+2)^2)/(49)=1 is (-4,3),(-4,-7)