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)

25 x^{2} + 9 y^{2} - 54 y - 144 = 0

\frac{( x + 3 ) ^{2}}{49} + \frac{( y - 1 ) ^{2}}{48} = 1

36 x^{2} + 9 y^{2} = 324

\frac{( x + 2 ) ^{2}}{100} + \frac{( y - 5 ) ^{2}}{36} = 1

25 x^{2} + 4 y^{2} - 100 x = 0

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))