What is the center ((x-5)^2)/4+((y+3)^2)/(16)=1 ?

The center ((x-5)^2)/4+((y+3)^2)/(16)=1 is (5,-3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

center ((x-5)^2)/4+((y+3)^2)/(16)=1

center

(5, - 3)

Solution steps

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

Rewrite

\frac{( x - 5 ) ^{2}}{4} + \frac{( y + 3 ) ^{2}}{16} = 1

in the form of the standard ellipse equation

\frac{( x - 5 ) ^{2}}{2 ^{2}} + \frac{( y - ( - 3 ) ) ^{2}}{4 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (5, - 3), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (5, - 3), b = 4, a = 2

And the center is:

(5, - 3)

What is the center ((x-5)^2)/4+((y+3)^2)/(16)=1 ?
The center ((x-5)^2)/4+((y+3)^2)/(16)=1 is (5,-3)