Problem 29 - Algebra and Trigonometry

Question

Complete the square to determine whether the equation represents an ellipse, a parabola, a hyperbola, or a degenerate conic. If the graph is an ellipse, find the center, foci, vertices, and lengths of the major and minor axes. If it is a parabola, find the vertex, focus, and directrix. If it is a hyperbola, find the center, foci, vertices, and asymptotes. Then sketch the graph of the equation. If the equation has no graph, explain why. $$ 3 x^{2}+4 y^{2}-6 x-24 y+39=0 $$

Step-by-Step Solution

Answer

The equation represents a degenerate conic as it forms a single point.

Step 1: Group x and y terms separately

Start by rearranging the given equation so that the terms involving \(x\) and \(y\) are grouped together: \[3 x^2 - 6 x + 4 y^2 - 24 y = -39\]

Step 2: Factor out coefficients of squared terms

Factor out the coefficients of the squared terms for each group:\[3(x^2 - 2x) + 4(y^2 - 6y) = -39\]

Step 3: Complete the square for x terms

To complete the square for \(x\), take half of the linear coefficient \(-2\), square it, and add and subtract it inside the parentheses:\[3((x^2 - 2x + 1) - 1)\]This becomes:\[3((x-1)^2 - 1)\]

Step 4: Complete the square for y terms

Similarly, complete the square for \(y\). Take half of \(-6\), square it, and add and subtract it:\[4((y^2 - 6y + 9) - 9)\]This becomes:\[4((y-3)^2 - 9)\]

Step 5: Rewrite the equation using completed squares

Substitute back the completed squares into the equation:\[3((x-1)^2 - 1) + 4((y-3)^2 - 9) = -39\]Simplify:\[3(x-1)^2 - 3 + 4(y-3)^2 - 36 = -39\]\[3(x-1)^2 + 4(y-3)^2 = 0\]

Step 6: Classify the conic

The resulting equation, \[3(x-1)^2 + 4(y-3)^2 = 0\], indicates a point conic because both terms are positive and their sum equals zero. This is a degenerate conic.

Key Concepts

Completing the Square

When solving conic sections, completing the square is a vital algebraic technique. It transforms quadratic equations into a more recognizable form, like a circle, ellipse, parabola, or hyperbola.
To complete the square, follow these steps:

• Identify and isolate the quadratic terms, making sure to group like terms together, such as those including the same variable.
• Next, factor any coefficients from the squared terms to simplify the process of completing the square.
• Add and subtract the same value within the equation to complete the square for that term. To find this value, take the coefficient of the linear term, divide by two, and then square it.

Completing the square not only helps identify the type of conic but also aids in finding elements like the center or vertex, depending on the conic form.

Ellipse

An ellipse is a closed curve formed by all points where the sum of the distances from two fixed points (the foci) is constant. This can be expressed in a standard equation form after completing the square for both the x and y terms.
The standard form is:\[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]

The Characteristics

• Center: Located at the point \(h, k\).
• Vertices: The endpoints of the longest and shortest diameters, along the axes.
• Major and Minor Axes: Length given by \(2a\) and \(2b\) respectively.
• Foci: Found at a distance \(c = \sqrt{a^2 - b^2}\) from the center along the major axis.

Ellipses are not present in the current problem since the sum of squares leads to zero.

Parabola

A parabola is the set of all points equidistant from a fixed point (the focus) and a line (the directrix). Parabolas typically appear in one of two forms, vertical or horizontal, depending on their orientation.
The general form is:\[(y-k)^2 = 4p(x-h)\] or \[(x-h)^2 = 4p(y-k)\]

Main Components

• Vertex: The point \(h, k\) where the parabola is at its peak.
• Focus: This lies at \(h, k + p\) for a vertical parabola, \(p\) units from the vertex.
• Directrix: The line \(y = k - p\) in a vertical parabola. It lies on the opposite side of the vertex from the focus.

Parabolas are not present in the current problem as both squared terms exist and the sum is not equal to zero as would be required for a parabola form.

Hyperbola

A hyperbola consists of two open curves or branches. All points on the hyperbola maintain a difference of distances to two fixed points, known as foci. The standard form for a hyperbola after completing the square is:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] or \[\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1\]

Key Elements

• Center: The midpoint between the foci, given by \(h, k\).
• Vertices: Located \(a\) units from the center.
• Foci: Situated \(c = \sqrt{a^2 + b^2}\) units from the center, along the transverse axis.
• Asymptotes: Do not intersect the hyperbola are given by the lines \(y = k \pm \frac{b}{a}(x-h)\).

The given problem does not indicate a hyperbola since the equation does not provide the standard hyperbola form due to the terms summing to zero.

Degenerate Conic

Degenerate conics occur in particular situations where conic sections don't form their usual curves. They might become points, lines, or intersecting lines depending on the conditions.
In our problem, we reach this conclusion based on the resulting equation:\[3(x-1)^2 + 4(y-3)^2 = 0\] Since the sum of completed square terms equals zero, this implies that both squared terms need to equal zero themselves, simultaneously. Therefore, both terms achieve zero only at \(x=1\) and \(y=3\), indicating this is a single point.

Understanding Degenerate Forms

• When equations simplify to zero, they generally indicate a degenerate form like:
• Point: When individual terms squared equal zero, like in this case.
• Line or Cross Lines: When the terms rearrange into linear equations.

Degenerate conics do not produce a graph with extended figures. Instead, they result in collapsed forms like a solitary point.