What is 2x^2+y^2-8y-2x+1=0 ?

The solution to 2x^2+y^2-8y-2x+1=0 is Ellipse with (h,k)=(1/2 ,4),a=(sqrt(31))/2 ,b=sqrt(31/2)

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2x^2+y^2-8y-2x+1=0

2 x^{2} + y^{2} - 8 y - 2 x + 1 = 0

(h, k) = (\frac{1}{2}, 4), a = \frac{31}{2}, b = \frac{31}{2}

Solution steps

2 x^{2} + y^{2} - 8 y - 2 x + 1 = 0

Rewrite

2 x^{2} + y^{2} - 8 y - 2 x + 1 = 0

in the form of the standard ellipse equation

\frac{( x - \frac{1}{2} ) ^{2}}{( \frac{31}{2} ) ^{2}} + \frac{( y - 4 ) ^{2}}{( \frac{31}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (\frac{1}{2}, 4), a = \frac{31}{2}, b = \frac{31}{2}

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (\frac{1}{2}, 4), b = \frac{31}{2}, a = \frac{31}{2}

8 x^{2} + 12 y^{2} = 96

\frac{x ^{2}}{9} + \frac{y ^{2}}{4} = \frac{x}{6} + \frac{y}{3}

\frac{( x + 1 ) ^{2}}{36} + \frac{( y + 2 ) ^{2}}{9} = 1

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

\frac{( x - 1.29 ) ^{2}}{1.17} + \frac{y ^{2}}{2.77} = 2

What is 2x^2+y^2-8y-2x+1=0 ?
The solution to 2x^2+y^2-8y-2x+1=0 is Ellipse with (h,k)=(1/2 ,4),a=(sqrt(31))/2 ,b=sqrt(31/2)