What is the foci ((x-3)^2}{16}+\frac{(y-2)^2)/9 =1 ?

The foci ((x-3)^2}{16}+\frac{(y-2)^2)/9 =1 is (3+sqrt(7),2),(3-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-3)^2}{16}+\frac{(y-2)^2)/9 =1

foci

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

(3 + 7, 2), (3 - 7, 2)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1 :

Ellipse with center

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

(3 + c, 2), (3 - c, 2)

Compute

c :

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

(3 + 7, 2), (3 - 7, 2)

x^{2} + 4 x + 4 y^{2} = 0

\frac{x ^{2}}{\frac{16}{9}} + \frac{y ^{2}}{16} = 1

(\frac{x}{5})^{2} + (\frac{y}{3})^{2} = 1

9 x^{2} + 16 y^{2} = 1

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

What is the foci ((x-3)^2}{16}+\frac{(y-2)^2)/9 =1 ?
The foci ((x-3)^2}{16}+\frac{(y-2)^2)/9 =1 is (3+sqrt(7),2),(3-sqrt(7),2)