Question

Identify the type of conic `4(x-2y+1)^2+9(2x+y+2)^2=25` ?

I reduced the equation to :
`40x^2+30xy+9y^2+60x+36y+15=0`
`a=40, b=9, h=18`
How to do the next steps?

Answer 1

It is an ellipse.

Explanation:

Let the equation be of the type `Ax^2+Bxy+Cy^2+Dx+Ey+F=0`

then if

`B^2-4AC=0` and `A=0` or `C=0`, it is a parabola

`B^2-4AC<0` and `A=C`, it is a circle

`B^2-4AC<0` and `A!=C`, it is an ellipse

`B^2-4AC>0`, it is a hyperbola

Now `4(x-2y+1)^2+9(2x+y+2)^2=25` can be written as

`4(x^2+4y^2+1-4xy+2x-4y)+9(4x^2+y^2+4+4xy+4y+8x)=25`

or `40x^2+20xy+25y^2+80x+20y+15=0`

and we have `A=40`, `B=20` and `C=25`

and hence `B^2-4AC=400-4000=-3600`

as `B^2-4AC < 0` and `A!=C`, it is an ellipse.

graph{4(x-2y+1)^2+9(2x+y+2)^2=25 [-3.219, 1.78, -1.23, 1.27]}

Note `if equation is`40x^2+30xy+9y^2+60x+36y+15=0#

we have `B^2-4AC=900-1440=-540` and as `A!=C`,

this is also an ellipse but different one.

graph{40x^2+30xy+9y^2+60x+36y+15=0 [-9.64, 10.36, -6.78, 3.22]}

Answer 2

See below.

Explanation:

Making

`{(u = x-2y+1),(v=2x+y+2):}`

which is equivalent to a coordinate's change so we have in the new coordinates

`4u^2+9v^2=25` or

`(u/(5/2))^2+(v/(5/3))^2=1` which is the equation of an ellipse.