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

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

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

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

(h, k) = (2, 1), a = 2, b = 22

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (2, 1), a = 2, b = 22

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, 1), b = 22, a = 2

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

\frac{X ^{2}}{27} + \frac{Y ^{2}}{36} = 1

x^{2} - 10 x + 4 y^{2} - 8 y = 5

\frac{( x + 3 ) ^{2}}{9} + \frac{( y - 2 ) ^{2}}{5} = 1

\frac{x ^{2}}{2} + \frac{y ^{2}}{3} = 1

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