What is x^2+2x+4y^2-16y+1=0 ?

The solution to x^2+2x+4y^2-16y+1=0 is Ellipse with (h,k)=(-1,2),a=4,b=2

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x^2+2x+4y^2-16y+1=0

x^{2} + 2 x + 4 y^{2} - 16 y + 1 = 0

(h, k) = (- 1, 2), a = 4, b = 2

Solution steps

x^{2} + 2 x + 4 y^{2} - 16 y + 1 = 0

Rewrite

x^{2} + 2 x + 4 y^{2} - 16 y + 1 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (- 1, 2), a = 4, b = 2

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

Ellipse with center (h, k) = (- 1, 2), a = 4, b = 2

\frac{( x - 1 ) ^{2}}{0. 1 ^{2}} + \frac{( y - 1 ) ^{2}}{0. 5 ^{2}} = 1

x^{2} + 2 y^{2} + 5 x - 2 y - 1 = 0

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

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

\frac{( x - 8 ) ^{2}}{12} + \frac{( y + 1 ) ^{2}}{16} = 1

What is x^2+2x+4y^2-16y+1=0 ?
The solution to x^2+2x+4y^2-16y+1=0 is Ellipse with (h,k)=(-1,2),a=4,b=2