# 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?

Jan 8, 2017

#### Explanation:

From the reference, the General Cartesian Form is:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0 \text{ [1]}$

Rewriting the given equation in the form:

$7 {x}^{2} + 0 x y + 4 {y}^{2} + - 28 x + 8 y + 4 = 0 \text{ [2]}$

We observe that $A = 7 , B = 0 \mathmr{and} C = 4$

Compute the discriminant:

${B}^{2} - 4 A C = {0}^{2} - 4 \left(7\right) \left(4\right) = - 112$

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:

${\left(y - k\right)}^{2} / {a}^{2} + {\left(x - h\right)}^{2} / {b}^{2} = 1 \text{ [3]}$

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

$7 {x}^{2} - 28 x + 7 {h}^{2} + 4 {y}^{2} + 8 y + 4 {k}^{2} = 7 {h}^{2} + 4 {k}^{2} - 4 \text{ [4]}$

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

$7 \left({x}^{2} - 4 x + {h}^{2}\right) + 4 \left({y}^{2} + 2 y + {k}^{2}\right) = 7 {h}^{2} + 4 {k}^{2} - 4 \text{ [4]}$

Use the expansions of

${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$

and

${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$

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

To find the value of h, use the equation:

$- 2 h x = - 4 x$

$h = 2$

To find the value of k, use the equation:

$- 2 k y = 2 y$

$k = - 1$

Substitute ${\left(x - 2\right)}^{2} \mathmr{and} {\left(y - - 1\right)}^{2}$ into equation [4]:

$7 {\left(x - 2\right)}^{2} + 4 {\left(y - - 1\right)}^{2} = 7 {h}^{2} + 4 {k}^{2} - 4 \text{ [5]}$

Substitute -1 for k and 2 for h:

$7 {\left(x - 2\right)}^{2} + 4 {\left(y - - 1\right)}^{2} = 7 {\left(2\right)}^{2} + 4 {\left(- 1\right)}^{2} - 4 = 28 \text{ [6]}$

Divide both sides of the equation by 28:

${\left(x - 2\right)}^{2} / 4 + {\left(y - - 1\right)}^{2} / 7 = 1 \text{ [7]}$

Write the denominators as squares:

${\left(x - 2\right)}^{2} / {2}^{2} + {\left(y - - 1\right)}^{2} / {\left(\sqrt{7}\right)}^{2} = 1 \text{ [8]}$

The center is $\left(2 , - 1\right)$
The major axis end points are $\left(2 , - 1 - \sqrt{7}\right) \mathmr{and} \left(2 , - 1 + \sqrt{7}\right)$
The minor axis end points are $\left(0 , - 1\right) \mathmr{and} \left(4 , - 1\right)$

Here is a graph: