Identifying Conics without Rotating Axes

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

$A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$

If we apply the rotation formulas to this equation we get the form

${A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime }{x}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime }{x}^{\prime }+{E}^{\prime }{y}^{\prime }+{F}^{\prime }=0$

It may be shown that ${B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }$. The expression does not vary after rotation, so we call the expression invariant. The discriminant, ${B}^{2}-4AC$, 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 $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ is transformed by rotating axes into the equation ${A}^{\prime }{{x}^{\prime }}^{2}+{B}^{\prime }{x}^{\prime }{y}^{\prime }+{C}^{\prime }{{y}^{\prime }}^{2}+{D}^{\prime }{x}^{\prime }+{E}^{\prime }{y}^{\prime }+{F}^{\prime }=0$, then ${B}^{2}-4AC={{B}^{\prime }}^{2}-4{A}^{\prime }{C}^{\prime }$.

The equation $A{x}^{2}+Bxy+C{y}^{2}+Dx+Ey+F=0$ is an ellipse, a parabola, or a hyperbola, or a degenerate case of one of these.

If the discriminant, ${B}^{2}-4AC$, 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.

1. $5{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0$
2. $5{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0$

Solution

1. Let’s begin by determining $A,B$, and $C$.
$\underset{A}{\underbrace{5}}{x}^{2}+\underset{B}{\underbrace{2\sqrt{3}}}xy+\underset{C}{\underbrace{2}}{y}^{2}-5=0$

Now, we find the discriminant.

$\begin{array}{l}{B}^{2}-4AC={\left(2\sqrt{3}\right)}^{2}-4\left(5\right)\left(2\right)\hfill \\ \text{ }=4\left(3\right)-40\hfill \\ \text{ }=12 - 40\hfill \\ \text{ }=-28<0\hfill \end{array}$

Therefore, $5{x}^{2}+2\sqrt{3}xy+2{y}^{2}-5=0$ represents an ellipse.

2. Again, let’s begin by determining $A,B$, and $C$.
$\underset{A}{\underbrace{5}}{x}^{2}+\underset{B}{\underbrace{2\sqrt{3}}}xy+\underset{C}{\underbrace{12}}{y}^{2}-5=0$

Now, we find the discriminant.

$\begin{array}{l}{B}^{2}-4AC={\left(2\sqrt{3}\right)}^{2}-4\left(5\right)\left(12\right)\hfill \\ \text{ }=4\left(3\right)-240\hfill \\ \text{ }=12 - 240\hfill \\ \text{ }=-228<0\hfill \end{array}$

Therefore, $5{x}^{2}+2\sqrt{3}xy+12{y}^{2}-5=0$ represents an ellipse.

Try It 3

Identify the conic for each of the following without rotating axes.

1. ${x}^{2}-9xy+3{y}^{2}-12=0$
2. $10{x}^{2}-9xy+4{y}^{2}-4=0$

Solution