Mastering Conic Sections Made EASY for the FE Exam!

In this article (and video above), we tackle a conic sections problem step by step, helping you determine the center of the curve and identify its type. By the end of this article, you’ll have the tools and confidence to handle similar questions with ease.

Question:

In this question, we are presented with the general equation of a conic section, and our job can be split up into two objectives: First, we need to determine the center of the conic section and secondly, we must identify the type of conic section it represents. Before diving into the solution, let’s take a moment to review some essential concepts that will help us approach this problem effectively.

Determine the center of the conic section described by the following general equation, and identify the type of conic section it represents:

9𝑥² + 16𝑦² − 54𝑥 + 64𝑦 = 311

Problem Context: Conic Sections

We start off by looking at what ‘conic sections’ refers to. This can be defined as any curve formed by the intersection of a plane with a right circular cone, as illustrated here. The type of curve—whether it’s an ellipse, parabola, circle, or hyperbola—depends on the angle at which the plane intersects the cone. Mathematically, we can distinguish between the different curves created by this intersecting plane using two key angles.

The first angle, denoted as θ (theta) in the FE Handbook, represents the angle between the intersecting plane and the vertical axis of the cone. The second angle, φ (phi), is the vertex angle, measured between the vertical axis of the cone, and its slanted outer surface. If these two angles are known, we can calculate what is referred to as the eccentricity of the conic sections using the formula:

This eccentricity can helps us classify the different curves. For example, an ellipse has an eccentricity of less than one (), while a parabola has an eccentricity equal to one (). Beyond this classification scheme, the FE Handbook also provides us with the standard equations for each of the conic sections. These equations not only define the shape of each curve but also provide us with key information, such as the center of each curve. The FE handbook defines this center as for each curve.

Step 1 – Rewrite the Equation in Standard Form

The equation given in this question is in its general form, but to find the center of the curve, we need to rewrite it in standard form in order for us to identify these h- and k-values. To start off, we group the x- and y-terms together, as shown here, dividing out any constants associated with the squared x- and y-terms.

Once grouped, we can proceed by completing the square for the x-terms. This involves taking the coefficient of the x-term, dividing it by two, and squaring the result. To maintain equivalence in the equation, this squared value is both added and subtracted within our brackets. This step ensures the expression can be factored into a perfect square trinomial, bringing us closer to the standard form.

9𝑥² + 16𝑦² − 54𝑥 + 64𝑦 = 311

1. Group x- and y-terms

9(𝑥² − 6𝑥) + 16(𝑦² + 4𝑦) = 311

2. Complete the square for x-terms

At this stage, we simplify squared terms to find that they equal 9. From here, the first three terms within the brackets can be easily factorized into , using either mental math or a calculator. However, it’s important to remember that the −9 term does not disappear—it forms part of the equation, ensuring we maintain equivalence to the original form.

9(𝑥² − 6𝑥 + (6/2)² − (6/2)²) + 16(𝑦² + 4𝑦) = 311

9(𝑥² − 6𝑥 + 9 − 9) + 16(𝑦² + 4𝑦) = 311

9((𝑥 − 3)² − 9) + 16(𝑦² + 4𝑦) = 311

3. Complete the square for y-terms

Next, we apply the same process to the y-terms. To complete the square, we take the coefficient of y, divide it by two, and then square the result. This squared value is added and subtracted within the brackets as before. And once this adjustment is made, the first three y-terms can be factorized into , with the subtracted constant remaining outside the factorized expression.

9((𝑥 − 3)² − 9) + 16(𝑦² + 4𝑦) = 311

9((𝑥 − 3)² − 9) + 16(𝑦² + 4𝑦 + (4/2)² − (4/2)²) = 311

9((𝑥 − 3)² − 9) + 16(𝑦² + 4𝑦 + 4 − 4) = 311

9((𝑥 − 3)² − 9) + 16((𝑦 + 2)² − 4) = 311

4. Simplify the equation

With the squares now completed, our next step is to simplify the equation. First, we multiply all of our terms inside the brackets with the constants outside to eliminate them. Then, we move all remaining constants to the right side of the equation, resulting in the simplified form shown here. When we compare this result to the standard forms of conic sections equations, we can clearly see that it matches the equation of an ellipse.

If we needed to determine the values of a and b in the standard form of the equation, we could divide the entire equation by the constant on the right-hand side to normalize it. However, since the problem only asks us to find the center of the curve, this additional step is unnecessary, since these values are already present within our brackets.

9((𝑥 − 3)² − 9) + 16((𝑦 + 2)² − 4) = 311

9(𝑥 − 3)² − 81 + 16(𝑦 + 2)² − 64 = 311

9(𝑥 − 3)² + 16(𝑦 + 2)² = 456

Step 2 – Identify the Center of the Conic Section

From the standard form of the equation, we know that the center corresponds to the h- and k-values. In this case, and . And since we’ve already confirmed that our equation represents an ellipse, we can now inspect the multiple-choice options, where we find that the correct answer is A.

It’s worth noting that the way these multiple-choice options are presented could easily cause confusion, especially if small mistakes are made. First, remember the distinction between the signs of h and k in the equation versus their coordinates. In other words, be careful not to misinterpret the center as by incorrectly swapping the signs. Second, the question indirectly tests our ability to distinguish between an ellipse and a hyperbola. The key difference lies in the sign between the two squared terms in the standard form—positive for an ellipse and negative for a hyperbola. When working quickly, it’s easy to confuse the two, so always double-check the equation to ensure that you accurately identify the curve, especially when using this method of visual inspection.

9(𝑥 − 3)² + 16(𝑦 + 2)² = 456

𝑐𝑒𝑛𝑡𝑒𝑟 = (ℎ, 𝑘)

Center: (3, −2)

Step 3 – Check your answer

Finally, it’s important to note that the FE Handbook provides specific conditions to help us differentiate between various conic sections using the original general equation. While this problem required us to rewrite the equation in standard form to find the center, for simpler questions where we only need to identify the type of curve, these conditions can offer a much quicker approach.

To demonstrate how these conditions work, let’s use the criteria for an ellipse to verify our answer. According to the general form provided in the FE Handbook, we identify the coefficients A, B, and C. In this equation, , as there is no xy-term. Additionally, and . Substituting these values into the condition , we find that the expression equals -576. Since this value is less than zero, it confirms that the curve is indeed an ellipse. This method provides a useful cross-check for visual inspection, ensuring that our answer is consistent and accurate.

9𝑥² + 16𝑦² − 54𝑥 + 64𝑦 = 311

(0)² − 4(9)(16) = −576

Ellipse!

FE Handbook:

The general form of the equation is: 𝐴𝑥² + 𝐵𝑥𝑦 + 𝐶𝑦² + 𝐷𝑥 + 𝐸𝑦 + 𝐹 = 0 where not both A and C are zero.

Ellipse: 𝐵² − 4𝐴𝐶 < 0

Hyperbola: 𝐵² − 4𝐴𝐶 > 0

Parabola: 𝐵² − 4𝐴𝐶 = 0

Circle: 𝐴 = 𝐶 and 𝐵 = 0

Straight line: 𝐴 = 𝐵 = 𝐶 = 0

The answer is A!

And that wraps up our problem! By carefully following the steps, we’ve determined with confidence that the correct answer is equal to A where the ellipse has a center of . This result aligns perfectly with both our analysis of the standard form and the conditions provided in the FE Handbook, confirming the accuracy of our solution.

Determine the center of the conic section described by the following general equation, and identify the type of conic section it represents:

9𝑥² + 16𝑦² − 54𝑥 + 64𝑦 = 311

Final Remarks:

To conclude, here are some key takeaways from this problem:

• Write equations in standard form: Be sure to practice transforming general expression to their standard form. The FE handbook provides us with a range of plug-and play formulas that help us identify key features, but in order to use them, you’d first need to identify the terms from this standard form.
• Use the conditions in the general form: For quicker identification of the type of curve, you can apply the conditions outlined in the FE Handbook directly to the general form of the equation. This can save valuable time in certain scenarios.
• Avoid mistakes with signs: Pay close attention to the signs in your equations, particularly when determining the center of a curve from its standard form. Small errors in signs can lead to incorrect results, especially when evaluating multiple-choice questions.
• Differentiate between standard equations visually: Familiarize yourself with the differences in the standard equations for each conic section. For example, recognize how the positive or negative signs between terms may distinguish an ellipse from a hyperbola.

By keeping these points in mind and practicing similar problems, you’ll easily develop the skills and confidence needed to tackle conic section questions efficiently and accurately.

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Anthony Fasano, P.E.
Engineering Management Institute
Author of Engineer Your Own Success