What is the foci 4x^2+y^2+6y+5=0 ?

The foci 4x^2+y^2+6y+5=0 is (0,-3+sqrt(3)),(0,-3-sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci 4x^2+y^2+6y+5=0

foci

4 x^{2} + y^{2} + 6 y + 5 = 0

(0, - 3 + 3), (0, - 3 - 3)

Solution steps

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

Calculate ellipse properties

4 x^{2} + y^{2} + 6 y + 5 = 0 :

Ellipse with center

(h, k) = (0, - 3), b = 2, a = 1

(0, - 3 + c), (0, - 3 - c)

Compute

c :

c = 2^{2} - 1^{2} : 3

(0, - 3 + 3), (0, - 3 - 3)

9 (x + 2)^{2} + (y + 1)^{2} = 1

\frac{( x - 1 ) ^{2}}{16} + \frac{( y + 2 ) ^{2}}{25} = 1

9 x^{2} + 25 y^{2} + 90 x = 0

\frac{x ^{2}}{8 ^{2}} + \frac{y ^{2}}{9 ^{2}} = 1

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

What is the foci 4x^2+y^2+6y+5=0 ?
The foci 4x^2+y^2+6y+5=0 is (0,-3+sqrt(3)),(0,-3-sqrt(3))