What is the vertices ((x-2)^2}{16}+\frac{y^2)/9 =1 ?

The vertices ((x-2)^2}{16}+\frac{y^2)/9 =1 is (6,0),(-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)^2}{16}+\frac{y^2)/9 =1

vertices

\frac{( x - 2 ) ^{2}}{16} + \frac{y ^{2}}{9} = 1

(6, 0), (- 2, 0)

Solution steps

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

Calculate ellipse properties

\frac{( x - 2 ) ^{2}}{16} + \frac{y ^{2}}{9} = 1 :

Ellipse with center

(h, k) = (2, 0), a = 4, b = 3

(2 + 4, 0), (2 - 4, 0)

Refine

(6, 0), (- 2, 0)

\frac{( x + 3 ) ^{2}}{36} + \frac{( y - 4 ) ^{2}}{25} = 1

\frac{( x ) ^{2}}{44} + \frac{y ^{2}}{125} = 1

\frac{( x - 5 ) ^{2}}{36} + \frac{( y + 6 ) ^{2}}{25} = 1

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

9 x^{2} + 4 y^{2} + 18 x - 8 y - 23 = 0

What is the vertices ((x-2)^2}{16}+\frac{y^2)/9 =1 ?
The vertices ((x-2)^2}{16}+\frac{y^2)/9 =1 is (6,0),(-2,0)