What is the foci ((x-1)^2)/4+((y+2)^2)/(16)=1 ?

The foci ((x-1)^2)/4+((y+2)^2)/(16)=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)/4+((y+2)^2)/(16)=1

foci

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{16} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{( x - 1 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{16} = 1 :

Ellipse with center

(h, k) = (1, - 2), b = 4, a = 2

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

Compute

c :

c = 4^{2} - 2^{2} : 23

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

9 x^{2} - 54 x + 4 y^{2} = - 45

\frac{( x + 2 ) ^{2}}{25} + \frac{( y - 1 ) ^{2}}{144} = 1

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

25 x^{2} + 9 y^{2} - 100 x + 18 y - 116 = 0

\frac{x ^{2}}{81} + \frac{y ^{2}}{49} = 1

What is the foci ((x-1)^2)/4+((y+2)^2)/(16)=1 ?
The foci ((x-1)^2)/4+((y+2)^2)/(16)=1 is (1,-2+2sqrt(3)),(1,-2-2sqrt(3))