What is the foci ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1 ?

The foci ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1 is (3+3sqrt(5),5),(3-3sqrt(5),5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1

foci

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1

(3 + 35, 5), (3 - 35, 5)

Solution steps

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

Calculate ellipse properties

\frac{( x - 3 ) ^{2}}{49} + \frac{( y - 5 ) ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (3, 5), a = 7, b = 2

(3 + c, 5), (3 - c, 5)

Compute

c :

c = 7^{2} - 2^{2} : 35

(3 + 35, 5), (3 - 35, 5)

\frac{( x - 1 ) ^{2}}{45} + \frac{( y - 2 ) ^{2}}{20} = 1

\frac{x ^{2}}{100} + \frac{y ^{2}}{16} = 1

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

\frac{x ^{2}}{1} + \frac{y ^{2}}{64} = 1

\frac{( x ) ^{2}}{13} + \frac{y ^{2}}{4} = 1

What is the foci ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1 ?
The foci ((x-3)^2}{49}+\frac{(y-5)^2)/4 =1 is (3+3sqrt(5),5),(3-3sqrt(5),5)