What is the foci 49x^2+25y^2-1225=0 ?

The foci 49x^2+25y^2-1225=0 is (0,2sqrt(6)),(0,-2sqrt(6))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci 49x^2+25y^2-1225=0

foci

49 x^{2} + 25 y^{2} - 1225 = 0

(0, 26), (0, - 26)

Solution steps

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

Calculate ellipse properties

49 x^{2} + 25 y^{2} - 1225 = 0 :

Ellipse with center

(h, k) = (0, 0), b = 7, a = 5

(0, 0 + c), (0, 0 - c)

Compute

c :

c = 7^{2} - 5^{2} : 26

(0, 0 + 26), (0, 0 - 26)

Refine

(0, 26), (0, - 26)

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

\frac{( x - 5 ) ^{2}}{9} + \frac{( y - 3 ) ^{2}}{16} = 1

\frac{( X - 1 ) ^{2}}{4} + \frac{( Y + 2 ) ^{2}}{16} = 1

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

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

What is the foci 49x^2+25y^2-1225=0 ?
The foci 49x^2+25y^2-1225=0 is (0,2sqrt(6)),(0,-2sqrt(6))