What is the foci ((x+4)^2}{16}+\frac{(y-2)^2)/9 =1 ?

The foci ((x+4)^2}{16}+\frac{(y-2)^2)/9 =1 is (-4+sqrt(7),2),(-4-sqrt(7),2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x+4)^2}{16}+\frac{(y-2)^2)/9 =1

foci

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

(- 4 + 7, 2), (- 4 - 7, 2)

Solution steps

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

Calculate ellipse properties

\frac{( x + 4 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (- 4, 2), a = 4, b = 3

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

Compute

c :

c = 4^{2} - 3^{2} : 7

(- 4 + 7, 2), (- 4 - 7, 2)

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

\frac{x ^{2}}{16} + \frac{y ^{2}}{49} = 1

\frac{( x + 5 ) ^{2}}{100} + \frac{( y - 4 ) ^{2}}{64} = 1

9 x^{2} + 16 y^{2} - 72 x + 64 y - 368 = 0

7 x^{2} + 2 y^{2} = 14

What is the foci ((x+4)^2}{16}+\frac{(y-2)^2)/9 =1 ?
The foci ((x+4)^2}{16}+\frac{(y-2)^2)/9 =1 is (-4+sqrt(7),2),(-4-sqrt(7),2)