What is the vertices 9x^2+y^2+18x+4y+4=0 ?

The vertices 9x^2+y^2+18x+4y+4=0 is (-1,1),(-1,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 9x^2+y^2+18x+4y+4=0

vertices

9 x^{2} + y^{2} + 18 x + 4 y + 4 = 0

(- 1, 1), (- 1, - 5)

Solution steps

(h, k + b), (h, k - b)

Calculate ellipse properties

9 x^{2} + y^{2} + 18 x + 4 y + 4 = 0 :

Ellipse with center

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

(- 1, - 2 + 3), (- 1, - 2 - 3)

Refine

(- 1, 1), (- 1, - 5)

\frac{x ^{2}}{9} + \frac{y ^{2}}{36} = 1

16 x^{2} + 25 y^{2} = 1600

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

16 x^{2} - 96 x + 9 y^{2} = 0

4 x^{2} + 25 y^{2} - 50 y = 75

What is the vertices 9x^2+y^2+18x+4y+4=0 ?
The vertices 9x^2+y^2+18x+4y+4=0 is (-1,1),(-1,-5)