What is the vertices (x^2)/9+(y^2)/5 =1 ?

The vertices (x^2)/9+(y^2)/5 =1 is (3,0),(-3,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2)/9+(y^2)/5 =1

vertices

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1

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

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{9} + \frac{y ^{2}}{5} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 3, b = 5

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

Refine

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

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

2 x^{2} + y^{2} = 50

9 x^{2} - 54 x + 4 y^{2} = - 45

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

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

What is the vertices (x^2)/9+(y^2)/5 =1 ?
The vertices (x^2)/9+(y^2)/5 =1 is (3,0),(-3,0)