What is 16x^2+4y^2-32x+24y-12=0 ?

The solution to 16x^2+4y^2-32x+24y-12=0 is Ellipse with (h,k)=(1,-3),a=2,b=4

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

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

(h, k) = (1, - 3), a = 2, b = 4

Solution steps

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

Rewrite

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

in the form of the standard ellipse equation

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

Therefore ellipse properties are:

(h, k) = (1, - 3), a = 2, b = 4

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (1, - 3), b = 4, a = 2

x^{2} - 2 x + \frac{y ^{2}}{4} - y \leq 14

x^{2} + 3 y^{2} = 6

2 x^{2} - 4 x + y^{2} - 6 y = 9

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

2 x^{2} + 3 y^{2} + 8 x + 8 y - 3 = 0

What is 16x^2+4y^2-32x+24y-12=0 ?
The solution to 16x^2+4y^2-32x+24y-12=0 is Ellipse with (h,k)=(1,-3),a=2,b=4