What is the center ((x+2)^2)/(16)+(y^2)/(49)=1 ?

The center ((x+2)^2)/(16)+(y^2)/(49)=1 is (-2,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center ((x+2)^2)/(16)+(y^2)/(49)=1

center

(- 2, 0)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - ( - 2 ) ) ^{2}}{4 ^{2}} + \frac{( y - 0 ) ^{2}}{7 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (- 2, 0), a = 4, b = 7

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (- 2, 0), b = 7, a = 4

And the center is:

(- 2, 0)

What is the center ((x+2)^2)/(16)+(y^2)/(49)=1 ?
The center ((x+2)^2)/(16)+(y^2)/(49)=1 is (-2,0)