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

The vertices (x^2)/4+(y^2)/3 =1 is (2,0),(-2,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2)/4+(y^2)/3 =1

vertices

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1

(2, 0), (- 2, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{4} + \frac{y ^{2}}{3} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 2, b = 3

(0 + 2, 0), (0 - 2, 0)

Refine

(2, 0), (- 2, 0)

9 x^{2} + 36 x + 9 y^{2} + 72 y + 155 = 0

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

x^{2} + 4 x + 4 y^{2} = 0

\frac{( x + 1 ) ^{2}}{36} + \frac{( y + 1 ) ^{2}}{64} = 1

4 x^{2} + y^{2} - 48 x - 4 y + 48 = 0

What is the vertices (x^2)/4+(y^2)/3 =1 ?
The vertices (x^2)/4+(y^2)/3 =1 is (2,0),(-2,0)