What is the vertices 9x^2+6y^2+36y=0 ?

The vertices 9x^2+6y^2+36y=0 is (0,0),(0,-6)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 9x^2+6y^2+36y=0

vertices

9 x^{2} + 6 y^{2} + 36 y = 0

(0, 0), (0, - 6)

Solution steps

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

Calculate ellipse properties

9 x^{2} + 6 y^{2} + 36 y = 0 :

Ellipse with center

(h, k) = (0, - 3), b = 3, a = 6

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

Refine

(0, 0), (0, - 6)

\frac{x ^{2}}{2} + \frac{y ^{2}}{3} = 1

x^{2} + 4 y^{2} - 6 x - 16 y = 36

(\frac{x + 5}{4})^{2} + (\frac{y - 4}{16})^{2} = 1

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

x^{2} + 9 y^{2} + 4 x = 5

What is the vertices 9x^2+6y^2+36y=0 ?
The vertices 9x^2+6y^2+36y=0 is (0,0),(0,-6)