Question

How do you identity if the equation `7x^2-28x+4y^2+8y=-4` is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Answer 1

It's an ellipse with `x`-semiaxis `2` and `y`-semiaxis `sqrt(7)` centred on the point `(2,-1)`

Explanation:

It is probably an ellipse because the coefficients of `x^2` and `y^2` have the same sign but different values, and there is no term in `xy`.

So complete the square for the `x` and `y` separately:
`7x^2-28x+4y^2+8y=-4`
`7(x-2)^2-28+4(y+1)^2-4=-4`
`7(x-2)^2+4(y+1)^2=28`
Then reduce to standard form:
`((x-2)/2)^2+((y+1)/sqrt(7))^2=1`
which is of the standard form:
`(x/a)^2+(y/b)^2=1`
so you can read off the semi-axes `a` and `b` and note that the centre is at `(2,-1)`.