Problem 30 - Precalculus: Mathematics for Calculus

Question

(a) Use the discriminant to identify the conic. (b) Confirm your answer by graphing the conic using a graphing device. $$x^{2}-2 x y+3 y^{2}=8$$

Step-by-Step Solution

Answer

The conic is an ellipse, confirmed by both discriminant and graphing.

Step 1: Identify the Coefficients

First, identify the coefficients in the given general quadratic equation \[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]For the given equation \[x^2 - 2xy + 3y^2 = 8\]we identify: \[A = 1, \, B = -2, \, C = 3, \, D = 0, \, E = 0, \, F = -8\]

Step 2: Calculate the Discriminant

The discriminant for a conic section is given by the formula \[\Delta = B^2 - 4AC\] Substitute the identified coefficients into this formula: \[(-2)^2 - 4 \times 1 \times 3 = 4 - 12 = -8\]

Step 3: Determine the Type of Conic

Use the value of the discriminant to determine the type of conic section:- If \(\Delta < 0\): Ellipse (and a circle if \(A = C\)).- If \(\Delta = 0\): Parabola.- If \(\Delta > 0\): Hyperbola.Since \(\Delta = -8\), the conic section is an ellipse.

Step 4: Graph the Conic Section

Using a graphing device, input the given equation\[x^2 - 2xy + 3y^2 = 8\]to plot it and visually confirm the shape. The graph should show an ellipse, confirming our analysis based on the discriminant.

Key Concepts

Discriminant

The discriminant helps us identify the type of conic section represented by a general quadratic equation. It's a simple computation that involves three key coefficients from the equation \[Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\]In this expression:

• The coefficient \(A\) is for \(x^2\).
• The coefficient \(B\) is for \(xy\).
• The coefficient \(C\) is for \(y^2\).

The discriminant formula for conic sections is given by:\[ \Delta = B^2 - 4AC \]Here's how it works:

• If \(\Delta < 0\), the conic section is an ellipse. This means it has a closed curve.
• If \(\Delta = 0\), it's a parabola, a curve that is open, extending infinitely in one direction.
• If \(\Delta > 0\), we have a hyperbola, which consists of two separate curves that mirror each other.

So, for the exercise, substituting the values into \(\Delta = B^2 - 4AC\), the result was \(-8\). Since \(\Delta < 0\), it confirms an ellipse.

Ellipse

An ellipse is a type of conic section that appears as an elongated circle or oval. It is defined by its geometric properties and is frequently encountered in physics and engineering.
The key idea is that for any point on the ellipse, the sum of the distances from two fixed points (known as foci) is constant.
The general equation of an ellipse can be derived from the standard quadratic form. When rearranged properly, it looks like:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]Here, \((h, k)\) are the coordinates of the center, while \(a\) and \(b\) are distances from this center to the edge along the x-axis and y-axis respectively.
However, in problems involving a general quadratic form like the one given, determining the ellipse's parameters requires identifying coefficients and analyzing the discriminant as done in the solution.

• If you graph the given quadratic equation from the exercise, it will indeed form an ellipse since \(\Delta < 0\).
• This is because the rotation of axes or skew terms (\(xy\) term) will influence how the equation forms on a graph.

General Quadratic Equation

The general quadratic equation is a versatile form that can represent various conic sections.It takes the form:\[ Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \]Understanding what each term represents is crucial:

• \(Ax^2\) and \(Cy^2\) are quadratic terms that describe basic shapes like parabolas and circles if solely present.
• \(Bxy\) introduces a rotation or distortion, typically changing circles to ellipses or hyperbolas.
• The linear terms \(Dx, Ey\) shift the conic's center across the coordinates.
• \(F\) is a constant that can further adjust position or represent degenerate forms when equal to zero.

This equation is powerful because, with the right coefficients, it can portray each classical conic determined by the discriminant \(\Delta\).
The exercise's quadratic equation gave us a glimpse of how these components interact to form an ellipse by computing the discriminant and confirming it through graphical means.