Let's rewrite equation so we can see what kind of conic it describes.

`2x^2+3y^2+4x-6y-13=0`

We can write this as

`2(x+1)^2-2+3(y-1)^2-3-13=0`

`2(x+1)^2+3(y-1)^2=18` now devide whole equation with 18

`(x+1)^2/9+(y-1)^2/6=1`

`((x+1)/3)^2+((y-1)/sqrt(6))^2=1`

And since equation of ellipse with center in `(p,q)` is `((x-p)/a)^2+((y-q)/b)^2=1` we can see that this is **ellipse** with center in `(-1,1)`.

We will use...

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Let's rewrite equation so we can see what kind of conic it describes.

`2x^2+3y^2+4x-6y-13=0`

We can write this as

`2(x+1)^2-2+3(y-1)^2-3-13=0`

`2(x+1)^2+3(y-1)^2=18` now devide whole equation with 18

`(x+1)^2/9+(y-1)^2/6=1`

`((x+1)/3)^2+((y-1)/sqrt(6))^2=1`

And since equation of ellipse with center in `(p,q)` is `((x-p)/a)^2+((y-q)/b)^2=1` we can see that this is **ellipse** with center in `(-1,1)`.

We will use the following formula to calculate distance from center to focus:

`f=sqrt(a^2-b^2)`

So in our case `f=sqrt(9-6)=sqrt3`.

**Foci:**

`F_1=(p-f,q)=(-1-sqrt3,1)`, `F_2=(p+f,q)=(-1+sqrt3,1)`

For **directrix** we use the following formula:

`x=p pm a^2/(sqrt(a^2-b^2))` sign `pm` is here because ellipse has 2 directrices.

So in our case: `x=-1pm 9/sqrt3=-1pm3sqrt3`

For more on conic sections see link below.