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How do we find the eccentricity of the conic #(3x+9)^2+(3y-12)^2-(2x-y)^2 = 6y-12x+9#?

Question

How do we find the eccentricity of the conic #(3x+9)^2+(3y-12)^2-(2x-y)^2 = 6y-12x+9#?
How do we find the eccentricity of the conic #(3x+9)^2+(3y-12)^2-(2x-y)^2 = 6y-12x+9#?

Answers

Eccentricity is #sqrt5/3#

Explanation:

Let us simplify #(3x+9)^2+(3y-12)^2-(2x-y)^2 = 6y-12x+9# and expand it.

#9x^2+54x+81+9y^2-72y+144-4x^2-y^2+4xy=6y-12x+9#

or #5x^2+8y^2+4xy+66x-78y+216=0#

The equation is of the type #Ax^2+Bxy+Cy^2+Dx+Ey+F=0# and as #B^2-4AC=16-160=-144<0# and #A!=C#,

it is an ellipse and #0 < e < 1#.

Its orientation is given by #theta=tan^(-1)((C-A+sqrt((C-A)^2+B^2))/B)#
= #tan^(-1)((8-5+sqrt((8-5)^2+4^2))/4)=tan^(-1)2#

and eccentricity is given by #e^2=(2sqrt(B^2+(A-C)^2))/(A+C+sqrt(B^2+(A-C)^2))#

= #(2sqrt(16+9))/(13+sqrt(16+9))=10/18=5/9#

and #e=sqrt5/3#

graph{(3x+9)^2+(3y-12)^2-(2x-y)^2 = 6y-12x+9 [-28.18, 11.82, -3.88, 16.12]}


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