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 [latex]{B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }[/latex]. The expression does not vary after rotation, so we call the expression invariant. The discriminant, [latex]{B}^{2}-4AC[/latex], 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 [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex] is transformed by rotating axes into the equation [latex]{A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime }{x}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime }{x}^{\prime }+{E}^{\prime }{y}^{\prime }+{F}^{\prime }=0[/latex], then [latex]{B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }[/latex].
The equation [latex]A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0[/latex] is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these.
If the discriminant, [latex]{B}^{2}-4AC[/latex], is
- [latex]<0[/latex], the conic section is an ellipse
- [latex]=0[/latex], the conic section is a parabola
- [latex]>0[/latex], 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.
- [latex]5{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0[/latex]
- [latex]5{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0[/latex]
Solution
- Let’s begin by determining [latex]A,B[/latex], and [latex]C[/latex].
[latex]\underset{A}{\underbrace{5}}{x}^{2}+\underset{B}{\underbrace{2\sqrt{3}}}xy+\underset{C}{\underbrace{2}}{y}^{2}-5=0[/latex]
Now, we find the discriminant.
[latex]\begin{array}{l}{B}^{2}-4AC={\left(2\sqrt{3}\right)}^{2}-4\left(5\right)\left(2\right)\hfill \\ \text{ }=4\left(3\right)-40\hfill \\ \text{ }=12 - 40\hfill \\ \text{ }=-28<0\hfill \end{array}[/latex]Therefore, [latex]5{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0[/latex] represents an ellipse.
- Again, let’s begin by determining [latex]A,B[/latex], and [latex]C[/latex].
[latex]\underset{A}{\underbrace{5}}{x}^{2}+\underset{B}{\underbrace{2\sqrt{3}}}xy+\underset{C}{\underbrace{12}}{y}^{2}-5=0[/latex]
Now, we find the discriminant.
[latex]\begin{array}{l}{B}^{2}-4AC={\left(2\sqrt{3}\right)}^{2}-4\left(5\right)\left(12\right)\hfill \\ \text{ }=4\left(3\right)-240\hfill \\ \text{ }=12 - 240\hfill \\ \text{ }=-228<0\hfill \end{array}[/latex]Therefore, [latex]5{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0[/latex] represents an ellipse.
Try It 3
Identify the conic for each of the following without rotating axes.
- [latex]{x}^{2}-9xy+3{y}^{2}-12=0[/latex]
- [latex]10{x}^{2}-9xy+4{y}^{2}-4=0[/latex]
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