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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

25 x^{2} + 16 y^{2} + 200 x + 64 y + 64 = 0

27 x^{2} + 48 y^{2} - 54 x - 192 y - 213 = 0

x^{2} + 16 y^{2} - 4 x - 128 y + 244 = 0

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

9 x^{(2)} + 16 y^{(2)} - 54 x + 64 y + 1 = 0

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