What is 4x^2+16y^2-4x-32y+1=0 ?

The solution to 4x^2+16y^2-4x-32y+1=0 is Ellipse with (h,k)=(1/2 ,1),a=2,b=1

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4x^2+16y^2-4x-32y+1=0

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0

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

Solution steps

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0

Rewrite

4 x^{2} + 16 y^{2} - 4 x - 32 y + 1 = 0

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

a > b

therefore

a

is semi-major axis and

b

is semi-minor axis

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

x = (y + 2)^{2}

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

f (x) = x^{2} + 5 x + 6

2 x^{2} + 2 y^{2} - 2 x - 2 y - 3 = 0

\frac{x ^{2}}{36} + \frac{y ^{2}}{25} = 1

What is 4x^2+16y^2-4x-32y+1=0 ?
The solution to 4x^2+16y^2-4x-32y+1=0 is Ellipse with (h,k)=(1/2 ,1),a=2,b=1