Question

How do you identity if the equation `5x^2+6x-4y=x^2-y^2-2x` is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Answer 1

It is an ellipse.

Explanation:

Let us write the equation in the form `Ax^2+Bxy+Cy^2+Dx+Ey+F=0`

As `5x^2+6x-4y=x^2-y^2-2x` is equivalent to

`5x^2-x^2+y^2+6x+2x-4y=0`

or `4x^2+y^2+8x-4y=0`

Hence, here we have `A=4`, `B=0` and `C=1`

and therefore `B^2-4AC=0-4xx4xx1=-16 < 0` and `A!=C`

Hence it is an ellipse (for details see here).

The equation can be written as `4(x^2+2x+1)+(y^2-4y+4)=8`

or `(x+1)^2/(1/4)+(y-2)^2/1=8`

or `(x+1)^2/2+(y-2)^2/8=1` and comparing it with general form of equation of an ellipse `(x-h)^2/a^2+(y-k)^2/b^2=1`, where center is `(h,k)` and axes of symmetry are `x=h` and `y=k`. The two axes are `2a` and `2b`.

Here, we have center at `(-1,2)` and axes of symmetry are `x=-1` and `y=2`. It has major axis `2sqrt8=4sqrt2` (vertically aligned as `b>a`) and minor axis is `2sqrt2`.

graph{5x^2+6x-4y=x^2-y^2-2x [-11.75, 8.25, -2.84, 7.16]}