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You have now seen that the equation Ax2 + By2 + Cx + Dy + F = 0 can represent: a circle when A and B are the same and non-zero. an ellipse when A and B are different numbers and the same sign. a hyperbola when A and B are different signs. a parabola when either A = 0 or B = 0 and the other coefficient is non-zero.

Sometimes, however, an equation will satisify one of the above conditions and will appear to describe one of the above conic sections, but it will turn out that there is really no solution, or an unexpected solution to the equation.

Let's look at an example of when this happens.

EXAMPLE 1
Determine what the equation 4x2 + 4y2 + 16x - 28y + 80 = 0 represents.
The equation 4x2 + 4y2 + 16x - 28y + 80 = 0 appears to represent a circle because the coefficients of x2 and y2 are the same and non-zero.
However, when you sketch the graph of this equation, we find that no graph exists. (Sketch this figure using the given action figure. )
Besides graphing the equation, we can also find out if 4x2 + 4y2 + 16x - 28y + 80 = 0 actually represents a circle by rewriting the equation into standard form. Recall how to do this in the following steps.
Step 1: Collect the x terms and the y terms together. (4x2 + 16x) + (4y2 - 28y) + 80 = 0 Step 2: Move any constant term to the right hand side of the equal sign. (4x2 + 16x) + (4y2 - 28y) = -80 Step 3: Factor the entire equation by the coefficient of x2 and y2. Cancel these factors. 4(x2 + 4x) + 4(y2 - 7y) = 4(-20) (x2 + 4x) + (y2 - 7y) = -20 Step 4: Find the numbers which complete each square. Add to both sides of the equation. (x2 + 4x + 4) + (y2 - 7y + 49/4) = -20 + 4 + 49/4 Step 5: Simplify the equation. (x + 2)2 + (y - 7/2)2 = -15/4
The equation (x + 2)2 + (y - 7/ 2)2 = -15/ 4 is now as close to the standard form (x - h)2 + (y - k)2 = r2 as we can get it. Do you see why it is not quite in standard form?
Notice that this equation cannot represent a circle because -15/4 < 0. In fact (x + 2)2 and (y - 7/2)2 are both positive for any values of x and y.
Because (x + 2)2 + (y - 7/2)2 = -15/4 has no solutions and this equation is equivalent to our original equation, we conclude that the solution set of 4x2 + 4y2 + 16x - 28y + 80 = 0 is empty. There is no curve!

In summary, you can guess what the graph of an equation of the form Ax2 + By2 + Cx + Dy + F = 0 might look like by looking at the coefficients.

However, since the equations can be degenerate in various ways, you need to rewrite the equation in standard form to check your guess.

In the following questions, you will practice identifying the graph of equations of the form Ax2 + By2 + Cx + Dy + F = 0.