What is the vertices 3x^2+7y^2-12x-28y+19=0 ?

The vertices 3x^2+7y^2-12x-28y+19=0 is (2+sqrt(7),2),(2-sqrt(7),2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 3x^2+7y^2-12x-28y+19=0

vertices

3 x^{2} + 7 y^{2} - 12 x - 28 y + 19 = 0

(2 + 7, 2), (2 - 7, 2)

Solution steps

(h + a, k), (h - a, k)

Calculate ellipse properties

3 x^{2} + 7 y^{2} - 12 x - 28 y + 19 = 0 :

Ellipse with center

(h, k) = (2, 2), a = 7, b = 3

(2 + 7, 2), (2 - 7, 2)

x^{2} + 3 y^{2} + 2 x - 12 y + 10 = 0

9 x^{2} + 16 y^{2} = 576

\frac{x ^{2}}{8} + \frac{y ^{2}}{5} = 1

9 x^{2} + 16 y^{2} - 144 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1

What is the vertices 3x^2+7y^2-12x-28y+19=0 ?
The vertices 3x^2+7y^2-12x-28y+19=0 is (2+sqrt(7),2),(2-sqrt(7),2)