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

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

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

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

(h, k) = (0, 2), a = 2, b = 2

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (0, 2), a = 2, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (0, 2), a = 2, b = 2

\frac{( x - 5 ) ^{2}}{25} + \frac{( y - 4 ) ^{2}}{9} = 1

9 x^{2} + 4 y^{2} + 18 x - 27 = 0

3 x^{2} + 4 y^{2} - 18 x - 8 y - 17 = 0

\frac{x ^{2}}{20} + \frac{y ^{2}}{16} = 1

\frac{( x + 7 ) ^{2}}{16} + \frac{( y - 4 ) ^{2}}{25} = 1

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