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

The solution to (2x)/(x^2+2y^2)=2 is Ellipse with (h,k)=(1/2 ,0),a= 1/2 ,b= 1/(2sqrt(2))

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(2x)/(x^2+2y^2)=2

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

(h, k) = (\frac{1}{2}, 0), a = \frac{1}{2}, b = \frac{1}{2 2}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

\frac{( x - \frac{1}{2} ) ^{2}}{( \frac{1}{2} ) ^{2}} + \frac{( y - 0 ) ^{2}}{( \frac{1}{2 2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (\frac{1}{2}, 0), a = \frac{1}{2}, b = \frac{1}{2 2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (\frac{1}{2}, 0), a = \frac{1}{2}, b = \frac{1}{2 2}

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- y^{2} - z^{2} = 1

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

6 x^{2} + 144 y^{2} = 864

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

What is (2x)/(x^2+2y^2)=2 ?
The solution to (2x)/(x^2+2y^2)=2 is Ellipse with (h,k)=(1/2 ,0),a= 1/2 ,b= 1/(2sqrt(2))