What is the foci ((y+4)^2)/(24)+((x-3)^2)/(12)=1 ?

The foci ((y+4)^2)/(24)+((x-3)^2)/(12)=1 is (3,-4+2sqrt(3)),(3,-4-2sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((y+4)^2)/(24)+((x-3)^2)/(12)=1

foci

\frac{( y + 4 ) ^{2}}{24} + \frac{( x - 3 ) ^{2}}{12} = 1

(3, - 4 + 23), (3, - 4 - 23)

Solution steps

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

Calculate ellipse properties

\frac{( y + 4 ) ^{2}}{24} + \frac{( x - 3 ) ^{2}}{12} = 1 :

Ellipse with center

(h, k) = (3, - 4), b = 26, a = 23

(3, - 4 + c), (3, - 4 - c)

Compute

c :

c = (26)^{2} - (23)^{2} : 23

(3, - 4 + 23), (3, - 4 - 23)

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

\frac{x ^{2}}{20} + \frac{y ^{2}}{16} = 1

\frac{y ^{2}}{16} + \frac{x ^{2}}{25} = 1

\frac{1}{4} (x - 5)^{2} + \frac{1}{9} (y + 5)^{2} = 1

\frac{( x + 1 ) ^{2}}{25} + \frac{( y + 1 ) ^{2}}{49} = 1

What is the foci ((y+4)^2)/(24)+((x-3)^2)/(12)=1 ?
The foci ((y+4)^2)/(24)+((x-3)^2)/(12)=1 is (3,-4+2sqrt(3)),(3,-4-2sqrt(3))