What is ((u+v)/(u^2+v^2))+((-v+u)/(u^2+v^2))=1 ?

The solution to ((u+v)/(u^2+v^2))+((-v+u)/(u^2+v^2))=1 is Ellipse with (h,k)=(1,0),a=1,b=1

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((u+v)/(u^2+v^2))+((-v+u)/(u^2+v^2))=1

(\frac{u + v}{u ^{2} + v ^{2}}) + (\frac{- v + u}{u ^{2} + v ^{2}}) = 1

(h, k) = (1, 0), a = 1, b = 1

Solution steps

\frac{u + v}{u ^{2} + v ^{2}} + \frac{- v + u}{u ^{2} + v ^{2}} = 1

Rewrite

\frac{u + v}{u ^{2} + v ^{2}} + \frac{- v + u}{u ^{2} + v ^{2}} = 1

in the form of the standard ellipse equation

\frac{( u - 1 ) ^{2}}{1 ^{2}} + \frac{( v - 0 ) ^{2}}{1 ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, 0), a = 1, b = 1

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, 0), b = 1, a = 1

3 x^{2} + 2 y^{2} - 6 = 0

\frac{y ^{2}}{64} + \frac{x ^{2}}{100} = 1

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{b} = 1

\frac{( x + 2 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

\frac{( x ^{(2)} )}{( 64 )} + \frac{( y ^{(2)} )}{( 36 )} = 1

What is ((u+v)/(u^2+v^2))+((-v+u)/(u^2+v^2))=1 ?
The solution to ((u+v)/(u^2+v^2))+((-v+u)/(u^2+v^2))=1 is Ellipse with (h,k)=(1,0),a=1,b=1