What is the vertices (x^2}{16}+\frac{y^2)/4 =1 ?

The vertices (x^2}{16}+\frac{y^2)/4 =1 is (4,0),(-4,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x^2}{16}+\frac{y^2)/4 =1

vertices

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1

(4, 0), (- 4, 0)

Solution steps

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

Calculate ellipse properties

\frac{x ^{2}}{16} + \frac{y ^{2}}{4} = 1 :

Ellipse with center

(h, k) = (0, 0), a = 4, b = 2

(0 + 4, 0), (0 - 4, 0)

Refine

(4, 0), (- 4, 0)

f (x) = x^{2} - 8 x + 12

\frac{y ^{2}}{36} - \frac{x ^{2}}{64} = 1

\frac{( x - 3 ) ^{2}}{5} + \frac{( y - 1 ) ^{2}}{15} = 1

3 x^{2} - y^{2} = 11

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

What is the vertices (x^2}{16}+\frac{y^2)/4 =1 ?
The vertices (x^2}{16}+\frac{y^2)/4 =1 is (4,0),(-4,0)