Problem 30 - A Graphical Approach to College Algebra

Question

Determine the type of conic section represented by each equation, and graph it, provided a graph exists. $$\frac{(x-4)^{2}}{8}+\frac{(y+1)^{2}}{2}=0$$

Step-by-Step Solution

Answer

The equation represents a degenerate conic, specifically a single point at (4, -1).

Step 1: Identify the Conic Section Equation Form

The given equation is \( \frac{(x-4)^{2}}{8} + \frac{(y+1)^{2}}{2} = 0 \). This resembles the standard form of an ellipse, \( \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \), but instead of "=1", this equation is equal to zero.

Step 2: Determine the Possibility of Real Solutions

For an ellipse, a parabola, or a hyperbola in a standard conic form, the equation should equal a non-zero value on the right side for it to represent a meaningful set of points in the Cartesian plane. Here, the equation equals zero, which restricts the possibilities.

Step 3: Analyze the Summands in the Equation

Each term in the equation is a square, \( \frac{(x-4)^{2}}{8} \geq 0 \) and \( \frac{(y+1)^{2}}{2} \geq 0 \). Since both non-negative terms sum to zero, each must independently equal zero. Thus, \( (x-4)^{2} = 0 \) and \( (y+1)^{2} = 0 \).

Step 4: Solve the Equations for x and y

Solve \( (x-4)^{2} = 0 \) leading to \( x = 4 \), and \( (y+1)^{2} = 0 \) leading to \( y = -1 \). Therefore, the set that satisfies both conditions is a single point: \((4, -1)\).

Step 5: Conclude the Type of Conic Section

Since the solution to the equation is a single point \((4, -1)\), the conic doesn't graph as a curve, but as a degenerate ellipse known as a 'point'. Therefore, the equation does not depict a conic section like a circle, ellipse, parabola, or hyperbola.

Key Concepts

Ellipse

An ellipse is a set of all points in a plane such that the sum of the distances from two fixed points (called the foci) is constant. Imagine stretching a circle along one direction that turns it into an ellipse.
The standard equation for an ellipse looks like this:

• Horizontal orientation: \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\)
• Vertical orientation: \(\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1\)

Here, \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, and \((h, k)\) is the center of the ellipse. Ellipses are closely related to circles, and when \(a = b\), it is essentially a circle. This geometric shape has applications in astronomy, as the orbits of planets are more elliptical than circular!
In the problem above, the equation doesn’t fit directly because it equals zero, not one. If it were equal to one, it might have represented a real ellipse centered at \((4, -1)\).

Degenerate Conic

A degenerate conic occurs when a conic section deteriorates into something simpler, such as a point, a line, or even two intersecting lines.
In terms of equations, this happens when certain parameters in the conic equation make it impossible for it to represent a complete curve.

• If all terms sum to zero, it can't describe a 'real' curve.
• In the exercise, the equation \(\frac{(x-4)^2}{8} + \frac{(y+1)^2}{2} = 0\) results in a single point \((4,-1)\). Both squares must equal zero since their sum equals zero.
• This single point indicates that we have a degenerate ellipse called a 'point'.

This degeneration effectively turns what could have been an elongated circular shape into just one location on the graph. Degenerations occur when equations satisfy minimal conditions like the sum being zero.

Graphing Conics

Graphing conics involves plotting quadratic equations that form different shapes like circles, ellipses, parabolas, and hyperbolas.
For graphing these shapes, understanding the equation's form and the particular parameters involved is key.

• The standard forms will guide you towards plotting the right curves.
• In the case of ellipses, identifying the center, axes lengths, and orientation can help sketch the shape.
• But not every conic equation can be graphed as a shape; some are degenerate and graph as simpler entities.

In our exercise, instead of graphing a curve, the conic is degenerate. Solving \((x-4)^2 = 0\) and \((y+1)^2 = 0\) shows us the graph is a single point at \((4, -1)\). This is a good example of why each conic doesn't always present as a curve. Degenerate conics, like the one in our problem, present unique graphing situations.