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

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

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

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

(h, k) = (- 1, 0), a = 1, b = \frac{1}{2}

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 0), a = 1, b = \frac{1}{2}

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 0), a = 1, b = \frac{1}{2}

\frac{x ^{2}}{2 5} + \frac{36 - 12 y + y ^{2}}{6} = 1

49 x^{2} + 100 y^{2} + 882 x - 1400 y + 3969 = 0

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

x^{2} + 4 y^{2} = 64

\frac{25 x ^{2} + 16 y ^{2}}{400} = 1

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