What is the vertices 5x^2+2y^2=5 ?

The vertices 5x^2+2y^2=5 is (0,sqrt(5/2)),(0,-sqrt(5/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 5x^2+2y^2=5

vertices

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

(0, \frac{5}{2}), (0, - \frac{5}{2})

Solution steps

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

Calculate ellipse properties

5 x^{2} + 2 y^{2} = 5 :

Ellipse with center

(h, k) = (0, 0), b = \frac{5}{2}, a = 1

(0, 0 + \frac{5}{2}), (0, 0 - \frac{5}{2})

Refine

(0, \frac{5}{2}), (0, - \frac{5}{2})

x^{2} + 3 y^{2} + 6 x - 6 y + 9 = 0

81 x^{2} + 16 y^{2} = 1296

25 x^{2} + 36 y^{2} - 350 x + 350 = 0

2 x^{2} - 8 x + 3 y^{2} + 6 y + 5 = 0

\frac{( x - 6 ) ^{2}}{49} + \frac{( y - 6 ) ^{2}}{16} = 1

What is the vertices 5x^2+2y^2=5 ?
The vertices 5x^2+2y^2=5 is (0,sqrt(5/2)),(0,-sqrt(5/2))