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Exercises Conic sections

Example 5 from Algebra and Trigonometry, 12.4 Rotation of Axes

Identifying the Conic without Rotating Axes

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

ⓐ $5 x^{2} + 2 \sqrt{3} x y + 2 y^{2} - 5 = 0$
ⓑ $5 x^{2} + 2 \sqrt{3} x y + 12 y^{2} - 5 = 0$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
- ⓐ Let’s begin by determining $A, B,$ and $C .$
  $\underset{A}{\underset{︸}{5}} x^{2} + \underset{B}{\underset{︸}{2 \sqrt{3}}} x y + \underset{C}{\underset{︸}{2}} y^{2} - 5 = 0$
  
  Now, we find the discriminant.
  
  $\begin{array}{l} B^{2} - 4 A C = {(2 \sqrt{3})}^{2} - 4 (5) (2) \\ = 4 (3) - 40 \\ = 12 - 40 \\ = - 28 < 0 \end{array}$
  Therefore, $5 x^{2} + 2 \sqrt{3} x y + 2 y^{2} - 5 = 0$ represents an ellipse.
- ⓑ Again, let’s begin by determining $A, B,$ and $C .$
  $\underset{A}{\underset{︸}{5}} x^{2} + \underset{B}{\underset{︸}{2 \sqrt{3}}} x y + \underset{C}{\underset{︸}{12}} y^{2} - 5 = 0$
  
  Now, we find the discriminant.
  
  $\begin{array}{l} B^{2} - 4 A C = {(2 \sqrt{3})}^{2} - 4 (5) (12) \\ = 4 (3) - 240 \\ = 12 - 240 \\ = - 228 < 0 \end{array}$
  Therefore, $5 x^{2} + 2 \sqrt{3} x y + 12 y^{2} - 5 = 0$ represents an ellipse.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

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

ⓐ $x^{2} - 9 x y + 3 y^{2} - 12 = 0$
ⓑ $10 x^{2} - 9 x y + 4 y^{2} - 4 = 0$

How did it go?

PreviousGraphing an Equation That Has No x′y′ Terms Next Identifying a Conic Given the Polar Form

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 12.4, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.