Problem 26 - 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. $$ 4 x^{2}-4 x-8 y+9=0 $$

Step-by-Step Solution

Answer

The equation represents a parabola with vertex \((\frac{1}{2}, \frac{5}{8})\), focus \((\frac{1}{2}, \frac{13}{8})\), and directrix \(y = \frac{1}{8}\).

Step 1: Rearrange the equation

First, we will rearrange the given equation to group the x and y terms. \[ 4x^2 - 4x - 8y + 9 = 0 \]Rewriting, we have:\[ 4(x^2 - x) - 8y + 9 = 0 \]

Step 2: Complete the square for x

To complete the square for the quadratic x term within the parentheses, we factor out a 4:\[ x^2 - x \]To complete the square, take half of the linear term coefficient (-1), square it, yielding 1/4, and adjust accordingly:\[ x^2 - x + \frac{1}{4} = \left(x - \frac{1}{2}\right)^2 \]Multiply this square by 4:\[ 4 \left( x^2 - x + \frac{1}{4} \right) = 4\left( x - \frac{1}{2} \right)^2 \]So, add and subtract \( \frac{1}{4} \times 4 = 1 \) inside the equation:\[ 4(x^2 - x + \frac{1}{4}) = 4(x - \frac{1}{2})^2 \]

Step 3: Adjust the equation

Now update the equation with the completed square term:\[ 4(x - \frac{1}{2})^2 - 4 - 8y + 9 = 0 \]Simplifying gives:\[ 4(x - \frac{1}{2})^2 - 8y + 5 = 0 \]

Step 4: Isolate y terms

Rearrange to solve for y and make it easier to identify the conic type:\[ -8y = -4(x - \frac{1}{2})^2 - 5 \]Divide every term by -8:\[ y = \frac{1}{2}(x - \frac{1}{2})^2 + \frac{5}{8} \]

Step 5: Identify the conic

The equation \( y = \frac{1}{2}(x - \frac{1}{2})^2 + \frac{5}{8} \) is in the standard form \( y = a(x - h)^2 + k \) of a parabola. Therefore, it represents a parabola.

Step 6: Find the vertex, focus, and directrix of the parabola

For the parabola \( y = \frac{1}{2}(x - \frac{1}{2})^2 + \frac{5}{8} \), the vertex \( (h, k) \) is \( (\frac{1}{2}, \frac{5}{8}) \).- \( a = \frac{1}{2} \), so the formula \( \frac{1}{4a} \) gives us the distance from the vertex to the focus or directrix.- \( \frac{1}{4a} = 2 \), so the focus is at \( (\frac{1}{2}, \frac{13}{8}) \).- The directrix is a horizontal line with equation \( y = \frac{1}{8} \).

Key Concepts

Completing the Square

Completing the square is a useful mathematical technique for rewriting quadratic equations. It transforms a quadratic expression into a perfect square trinomial, making it easier to manipulate or graph the equation. This process is particularly important when working with conic sections such as parabolas, circles, and ellipses.
To complete the square, follow these general steps:

• Start with a quadratic equation of the form: \( ax^2 + bx + c \).
• Rearrange terms to group \( x \) and \( y \) terms separately.
• Factor out any coefficients from the quadratic term if necessary.
• Take half of the coefficient of the linear term \( x \), square it, and add this square into and subtract it from the equation.
• Simplify to express the equation as a perfect square.

In the given problem, we applied completing the square to the \( x \) terms in the equation \( 4x^2 - 4x - 8y + 9 = 0 \). This allowed us to identify the conic section involved, which was a crucial step in solving the problem.

Parabola

A parabola is a symmetric, U-shaped curve that can open upwards, downwards, left, or right. It is one of the simplest forms of conic sections and can be represented in a few different ways including the standard form. In a parabola, every point is equidistant from a single point called the focus and a line called the directrix.
We identified the given equation \( y = \frac{1}{2}(x - \frac{1}{2})^2 + \frac{5}{8} \) as a parabola. This is a classic example of the standard form of a parabola along its vertical axis. The pattern generally looks like \( y = a(x - h)^2 + k \), where \((h, k)\) is the vertex. Understanding the orientation and shape of a parabola can help predict the path of objects or rays of light, explain the trajectory of projectiles, and in many other applications.

Vertex of a Parabola

The vertex of a parabola is its maximum or minimum point, where the parabola changes direction. In the vertex form of a parabola's equation, \( y = a(x - h)^2 + k \), the vertex is represented by the coordinates \((h, k)\). This single point provides vital information about the entire curve.
For the given parabola \( y = \frac{1}{2}(x - \frac{1}{2})^2 + \frac{5}{8} \), the vertex is located at \( (\frac{1}{2}, \frac{5}{8}) \). Since \( a > 0 \), the parabola opens upwards and the vertex is the lowest point. Knowing the vertex makes it easier to graph the parabola and find its other features like the focus and directrix.

Focus of a Parabola

The focus of a parabola is a point from which distances to any point on the parabola are precisely related to their corresponding distances to the directrix (a line not touching the parabola). In simpler terms, it is a key characteristic that helps define the shape and path of a parabola.
To find the focus, use the formula: \( \frac{1}{4a} \) to determine the distance between the vertex and the focus along the axis of symmetry. For the parabola \( y = \frac{1}{2}(x - \frac{1}{2})^2 + \frac{5}{8} \), we calculated this distance as 2, resulting in the focus at the point \((\frac{1}{2}, \frac{13}{8})\). By understanding the location of the focus, you can gain insights into how the parabola is oriented in space, and it is crucial for applications involving reflective properties, like satellite dishes and headlights.