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

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

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

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

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

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

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

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

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

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

\frac{( x - 5 ) ^{2}}{49} + \frac{( y + 4 ) ^{2}}{81} = 1

x^{2} + 9 y^{2} \leq 16

(\frac{x - 2}{2})^{2} + (\frac{y + 3}{4})^{2} = 1

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

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