# Determine wether `2x^2+3y^2+4x-6y-13=0`  is a parabola, ellipse, or hyperbola. Give the center, foci, directrix...etc Please show how to differentiate between the conic sections in this problem and show how to solve. 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.

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