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

Please open this link General Cartesian Form in another tab and read, carefully.

Explanation:

From the reference, the General Cartesian Form is:

`Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0" [1]"`

Rewriting the given equation in the form:

`7x^2 + 0xy + 4y^2 + -28x + 8y + 4 = 0" [2]"`

We observe that `A = 7, B = 0 and C = 4`

Compute the discriminant:

`B^2 - 4AC = 0^2 - 4(7)(4) = -112`

According to the conditions described in the reference, the discriminant is less than zero, therefore, the equation describes an ellipse.

Because `B = 0`, the ellipse is not rotated from either a horizontal or vertical major axis. Because `C < A`, the ellipse is has a vertical major axis. Therefore, I recommend that we convert the equation into the standard form:

`(y - k)^2/a^2 + (x - h)^2/b^2 = 1" [3]"`

Because `A = 7 and C = 4`, add `7h^2 + 4k^2` to both sides of equation [2] and move the constant back to the right side:

`7x^2-28x + 7h^2 + 4y^2 + 8y + 4k^2 = 7h^2 +4k^2 -4" [4]"`

Factor a 7 out of the first 3 term and a 4 out of the next 3 terms:

`7(x^2-4x + h^2) + 4(y^2 + 2y + k^2) = 7h^2 +4k^2 -4" [4]"`

Use the expansions of

`(x - h)^2 = x^2 - 2hx + h^2`

and

`(y - k)^2 = y^2 - 2ky + k^2`

to match the middle terms with the middle terms in equation [4]:

To find the value of h, use the equation:

`-2hx = -4x`

`h = 2`

To find the value of k, use the equation:

`-2ky = 2y`

`k = -1`

Substitute `(x - 2)^2 and (y - -1)^2` into equation [4]:

`7(x - 2)^2 + 4(y - -1)^2 = 7h^2 +4k^2 -4" [5]"`

Substitute -1 for k and 2 for h:

`7(x - 2)^2 + 4(y - -1)^2 = 7(2)^2 +4(-1)^2 -4 = 28" [6]"`

Divide both sides of the equation by 28:

`(x - 2)^2/4 + (y - -1)^2/7 = 1" [7]"`

Write the denominators as squares:

`(x - 2)^2/2^2 + (y - -1)^2/(sqrt(7))^2 = 1" [8]"`

The center is `(2, -1)`
The major axis end points are `(2, -1-sqrt(7)) and (2, -1+sqrt(7))`
The minor axis end points are `(0, -1) and (4, -1)`

Here is a graph: