What is 3x^2-12x+y^2-4y+1=0 ?

The solution to 3x^2-12x+y^2-4y+1=0 is Ellipse with (h,k)=(2,2),a=sqrt(5),b=sqrt(15)

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3x^2-12x+y^2-4y+1=0

3 x^{2} - 12 x + y^{2} - 4 y + 1 = 0

(h, k) = (2, 2), a = 5, b = 15

Solution steps

3 x^{2} - 12 x + y^{2} - 4 y + 1 = 0

Rewrite

3 x^{2} - 12 x + y^{2} - 4 y + 1 = 0

in the form of the standard ellipse equation

\frac{( x - 2 ) ^{2}}{( 5 ) ^{2}} + \frac{( y - 2 ) ^{2}}{( 15 ) ^{2}} = 1

Therefore ellipse properties are:

(h, k) = (2, 2), a = 5, b = 15

b > a

therefore

b

is semi-major axis and

a

is semi-minor axis

Ellipse with center (h, k) = (2, 2), b = 15, a = 5

(\frac{u + v}{u ^{2} + v ^{2}}) + (\frac{- v + u}{u ^{2} + v ^{2}}) = 1

3 x^{2} + 2 y^{2} - 6 = 0

\frac{y ^{2}}{64} + \frac{x ^{2}}{100} = 1

\frac{( x + 2 ) ^{2}}{64} + \frac{( y - 3 ) ^{2}}{b} = 1

\frac{( x + 2 ) ^{2}}{4} + \frac{( y + 2 ) ^{2}}{9} = 1

What is 3x^2-12x+y^2-4y+1=0 ?
The solution to 3x^2-12x+y^2-4y+1=0 is Ellipse with (h,k)=(2,2),a=sqrt(5),b=sqrt(15)