What is x^2+2x+4y^2-8y+1=0 ?

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

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x^2+2x+4y^2-8y+1=0

x^{2} + 2 x + 4 y^{2} - 8 y + 1 = 0

(h, k) = (- 1, 1), a = 2, b = 1

Solution steps

x^{2} + 2 x + 4 y^{2} - 8 y + 1 = 0

Rewrite

x^{2} + 2 x + 4 y^{2} - 8 y + 1 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 1), a = 2, b = 1

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 1), a = 2, b = 1

25 x^{2} + 4 y^{2} - 8 y - 50 x = 71

3 y^{2} = - 2 x^{2} + 1

12 x^{2} + 16 y^{2} = 192

\frac{x ^{2}}{c _{1}^{2}} = 1 - \frac{y ^{2}}{c _{2}^{2}}

16 x^{2} + 4 y^{2} - 32 x + 24 y - 12 = 0

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