Now we have come full circle. How do we identify the type of conic described by an equation? What happens when the axes are rotated? Recall, the general form of a conic is
If we apply the rotation formulas to this equation we get the form
It may be shown that B2−4AC=B′2−4A′C′. The expression does not vary after rotation, so we call the expression invariant. The discriminant, B2−4AC, is invariant and remains unchanged after rotation. Because the discriminant remains unchanged, observing the discriminant enables us to identify the conic section.
A General Note: Using the Discriminant to Identify a Conic
If the equation Ax2+Bxy+Cy2+Dx+Ey+F=0 is transformed by rotating axes into the equation A′x′2+B′x′y′+C′y′2+D′x′+E′y′+F′=0, then B2−4AC=B′2−4A′C′.
The equation Ax2+Bxy+Cy2+Dx+Ey+F=0 is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these.
If the discriminant, B2−4AC, is
- <0, the conic section is an ellipse
- =0, the conic section is a parabola
- >0, the conic section is a hyperbola
Example 5: Identifying the Conic without Rotating Axes
Identify the conic for each of the following without rotating axes.
- 5x2+2√3xy+2y2−5=0
- 5x2+2√3xy+12y2−5=0
Solution
- Let’s begin by determining A,B, and C.
5⏟Ax2+2√3⏟Bxy+2⏟Cy2−5=0
Now, we find the discriminant.
B2−4AC=(2√3)2−4(5)(2) =4(3)−40 =12−40 =−28<0Therefore, 5x2+2√3xy+2y2−5=0 represents an ellipse.
- Again, let’s begin by determining A,B, and C.
5⏟Ax2+2√3⏟Bxy+12⏟Cy2−5=0
Now, we find the discriminant.
B2−4AC=(2√3)2−4(5)(12) =4(3)−240 =12−240 =−228<0Therefore, 5x2+2√3xy+12y2−5=0 represents an ellipse.
Try It 3
Identify the conic for each of the following without rotating axes.
- x2−9xy+3y2−12=0
- 10x2−9xy+4y2−4=0
Candela Citations
- Precalculus. Authored by: OpenStax College. Provided by: OpenStax. Located at: http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. License: CC BY: Attribution