What is ((x-2)^2)/4+((y+2)^2)/8 =1 ?

The solution to ((x-2)^2)/4+((y+2)^2)/8 =1 is Ellipse with (h,k)=(2,-2),a=2,b=2sqrt(2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

((x-2)^2)/4+((y+2)^2)/8 =1

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{8} = 1

(h, k) = (2, - 2), a = 2, b = 22

Solution steps

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{8} = 1

Rewrite

\frac{( x - 2 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{8} = 1

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, - 2), a = 2, b = 22

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, - 2), b = 22, a = 2

What is ((x-2)^2)/4+((y+2)^2)/8 =1 ?
The solution to ((x-2)^2)/4+((y+2)^2)/8 =1 is Ellipse with (h,k)=(2,-2),a=2,b=2sqrt(2)