What is the vertices 12(x+4)^2+3(y-1)^2=48 ?

The vertices 12(x+4)^2+3(y-1)^2=48 is (-4,5),(-4,-3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 12(x+4)^2+3(y-1)^2=48

vertices

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48

(- 4, 5), (- 4, - 3)

Solution steps

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

Calculate ellipse properties

12 (x + 4)^{2} + 3 (y - 1)^{2} = 48 :

Ellipse with center

(h, k) = (- 4, 1), b = 4, a = 2

(- 4, 1 + 4), (- 4, 1 - 4)

Refine

(- 4, 5), (- 4, - 3)

4 x^{2} + 25 y^{2} = 100

9 x^{2} + 18 x + 4 y^{2} - 24 y + 9 = 0

3 x^{2} + 12 y^{2} - 6 x - 48 y + 33 = 0

36 x^{2} + 25 y^{2} = 900

x^{2} + 2 y^{2} + 2 x - 20 y + 43 = 0

What is the vertices 12(x+4)^2+3(y-1)^2=48 ?
The vertices 12(x+4)^2+3(y-1)^2=48 is (-4,5),(-4,-3)