What is the center ((x-6)^2)/5+((y+1)^2)/6 =1 ?

The center ((x-6)^2)/5+((y+1)^2)/6 =1 is (6,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center ((x-6)^2)/5+((y+1)^2)/6 =1

center

(6, - 1)

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (6, - 1), a = 5, b = 6

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (6, - 1), b = 6, a = 5

And the center is:

(6, - 1)

What is the center ((x-6)^2)/5+((y+1)^2)/6 =1 ?
The center ((x-6)^2)/5+((y+1)^2)/6 =1 is (6,-1)