What is the foci 9x^2+4y^2+54x-8y+49=0 ?

The foci 9x^2+4y^2+54x-8y+49=0 is (-3,1+sqrt(5)),(-3,1-sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

foci 9x^2+4y^2+54x-8y+49=0

foci

9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0

(- 3, 1 + 5), (- 3, 1 - 5)

Solution steps

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

Calculate ellipse properties

9 x^{2} + 4 y^{2} + 54 x - 8 y + 49 = 0 :

Ellipse with center

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

(- 3, 1 + c), (- 3, 1 - c)

Compute

c :

c = 3^{2} - 2^{2} : 5

(- 3, 1 + 5), (- 3, 1 - 5)

25 (x - 1)^{2} + 9 (y + 4)^{2} = 225

\frac{( x + 1 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{9} = 1

\frac{( x + 2 ) ^{2}}{16} + \frac{( y - 3 ) ^{2}}{49} = 1

\frac{( x - 6 ) ^{2}}{9} + \frac{( y - 9 ) ^{2}}{64} = 1

y^{2} = 16 - x^{2}

What is the foci 9x^2+4y^2+54x-8y+49=0 ?
The foci 9x^2+4y^2+54x-8y+49=0 is (-3,1+sqrt(5)),(-3,1-sqrt(5))