# How do you identity if the equation 2x^2+12x+18-y^2=3(2-y^2)+4y is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Jan 1, 2017

Rewrite the equation in General Cartesian Form and then use the conditions of the discriminant to make the determination.

#### Explanation:

From the reference Conic Sections - General Cartesian Form , the form is:

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

Rewrite the given equation to match this form:

$2 {x}^{2} + 0 x y + 2 {y}^{2} + 12 x - 4 y + 12 = 0 \text{ [2]}$

$A = 2 , B = 0 , \mathmr{and} C = 2$

The discriminant is:

${B}^{2} - 4 A C = - 16$

The conditions say that, when the discriminant is less than zero, then the equation represents and ellipse but, when $A = C$, the equation is a special case of an ellipse; a circle. Therefore, this equation is a circle and it must fit the standard Cartesian form for a circle:

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

where x and y correspond to any point $\left(x , y\right)$ on the circle, h and k correspond to the center point $\left(h , k\right)$, and r is the radius.

Expand the squares of equation [3]:

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

To make equation [2] look more like equation [4] we divide both sides by 2 and move the constant term to the right side:

${x}^{2} + {y}^{2} + 6 x - 2 y = - 6 \text{ [5]}$

Add ${h}^{2} \mathmr{and} {k}^{2}$ to both sides of the equation:

${x}^{2} + 6 x + {h}^{2} + {y}^{2} - 2 y + {k}^{2} = {h}^{2} + {k}^{2} - 6 \text{ [6]}$

We can find the value of h is we set the second term in equation [4] equal to the second term in equation [6] equal:

$- 2 h x = 6 x$

$h = - 3$

We can do the same for k with the fifth terms in equations [4] and [6]:

$- 2 k x = - 2 x$

$k = 1$

Now that we know the values of h and k, we can substitute the squares into equation [6] and compute the values of ${h}^{2} \mathmr{and} {k}^{2}$ on the right:

${\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {\left(- 3\right)}^{2} + {1}^{2} - 6 = 9 + 1 - 6 = 4 \text{ [7]}$

Write the right side as a square:

${\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = {2}^{2} \text{ [8]}$

To graph equation [8], set your compass to a radius of 2, make the center point $\left(- 3 , 1\right)$ and draw a circle.

Here is a graph: