What is x^2+2y^2-2x-4y-8=0 ?

The solution to x^2+2y^2-2x-4y-8=0 is Ellipse with (h,k)=(1,1),a=sqrt(11),b=sqrt(11/2)

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x^2+2y^2-2x-4y-8=0

x^{2} + 2 y^{2} - 2 x - 4 y - 8 = 0

(h, k) = (1, 1), a = 11, b = \frac{11}{2}

Solution steps

x^{2} + 2 y^{2} - 2 x - 4 y - 8 = 0

Rewrite

x^{2} + 2 y^{2} - 2 x - 4 y - 8 = 0

in the form of the standard ellipse equation

\frac{( x - 1 ) ^{2}}{( 11 ) ^{2}} + \frac{( y - 1 ) ^{2}}{( \frac{11}{2} ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (1, 1), a = 11, b = \frac{11}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (1, 1), a = 11, b = \frac{11}{2}

7 x^{2} + 25 y^{2} = 175

\frac{( x - 1 ) ^{2}}{16} + \frac{( y - 2 ) ^{2}}{9} = 1

x^{2} + 5 y^{2} = 9 (2.1)

81 x^{2} + 16 y^{2} = 1296

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

What is x^2+2y^2-2x-4y-8=0 ?
The solution to x^2+2y^2-2x-4y-8=0 is Ellipse with (h,k)=(1,1),a=sqrt(11),b=sqrt(11/2)