Question

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

Answer 1

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]}