Problem 26 - Precalculus: Mathematical for Calculus

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 describes a parabola. Vertex: \(\left(\frac{1}{2}, 1\right)\), Focus: \(\left(\frac{1}{2}, \frac{3}{2}\right)\), Directrix: \(y = \frac{1}{2}\).

Step 1: Re-arrange and Group Terms

Start by moving all terms to one side of the equation and group the terms with the same variable. The given equation is:\[ 4x^2 - 4x - 8y + 9 = 0 \]Re-arrange the equation as follows:\[ 4x^2 - 4x = 8y - 9 \]

Step 2: Complete the Square for x

To complete the square for the terms involving \(x\), factor out the coefficient of \(x^2\), which is 4:\[ 4(x^2 - x) = 8y - 9 \]To complete the square inside the parenthesis, take half of the linear coefficient (which is -1), square it (\((-\frac{1}{2})^2 = \frac{1}{4}\)), and add inside the parenthesis:\[ 4\left(x^2 - x + \frac{1}{4}\right) - 4 \cdot \frac{1}{4} = 8y - 9 \]This modifies to:\[ 4(x - \frac{1}{2})^2 - 1 = 8y - 9 \]

Step 3: Simplify and Solve for y

Add 1 to both sides to keep the equation balanced:\[ 4(x - \frac{1}{2})^2 = 8y - 8 \]Divide every term by 8 to solve for \(y\):\[ \frac{1}{2}(x - \frac{1}{2})^2 = y - 1 \]

Step 4: Identify the Conic Section

The equation is now \( y = \frac{1}{2}(x - \frac{1}{2})^2 + 1 \), which is in the vertex form of a parabola \( y = a(x-h)^2 + k \).From this, identify the vertex (h, k) as \( \left(\frac{1}{2}, 1\right) \). The parabola opens upwards since \(a = \frac{1}{2} > 0\).

Step 5: Find the Focus and Directrix of the Parabola

For a parabola given by \( y = a(x-h)^2 + k \), the focus is at \( \left(h, k + \frac{1}{4a}\right) \) and the directrix is \( y = k - \frac{1}{4a} \).Here, \( a = \frac{1}{2} \), \( h = \frac{1}{2} \), \( k = 1 \).Focus: \( \left(\frac{1}{2}, 1 + \frac{1}{2}\right) = \left(\frac{1}{2}, \frac{3}{2}\right) \).Directrix: \( y = 1 - \frac{1}{2} = \frac{1}{2} \).

Step 6: Sketch the Parabola

With the vertex at \( \left(\frac{1}{2}, 1\right) \), the focus at \( \left(\frac{1}{2}, \frac{3}{2}\right) \), and the directrix \( y = \frac{1}{2} \), you can sketch the parabola opening upwards. The parabola is symmetric around the vertical line \( x = \frac{1}{2} \).

Key Concepts

Completing the Square

Completing the square is a technique used to convert a quadratic expression into a perfect square trinomial. This method is especially useful in solving quadratic equations and analyzing conic sections, such as parabolas. The idea is to manipulate the expression into the form

• \( (x-h)^2 = k \) for some values \(h\) and \(k\).

Let's go through this process with an example from the exercise:Suppose we have the equation \(4x^2 - 4x - 8y + 9 = 0\). Our first step is to rearrange the terms involving \(x\):

• Rewrite the equation as \(4x^2 - 4x = 8y - 9\).
• Factor out the 4 from \(4x^2 - 4x\), which results in \(4(x^2 - x)\).

Next, we complete the square on \(x^2 - x\):

• Take half of the coefficient of \(x\), which is \(-1\), divide it by two and square it to get \((\frac{-1}{2})^2 = \frac{1}{4}\).
• Add and subtract \(\frac{1}{4}\) inside the parentheses: \(4\left(x^2 - x + \frac{1}{4}\right) - 4 \cdot \frac{1}{4} = 8y - 9\).

This gives us a perfect square trinomial: \(4(x - \frac{1}{2})^2 - 1 = 8y - 9\).This transformation is key in identifying the characteristics of the quadratic, such as its graph and vertex location.

Parabola

A parabola is a U-shaped curve described by a quadratic equation. The standard vertex form of a parabola is:

• \( y = a(x-h)^2 + k \).

where \( (h, k) \) is the vertex of the parabola, and \(a\) determines the width and direction of opening. In our exercise, we've manipulated the equation to:

• \( y = \frac{1}{2}(x-\frac{1}{2})^2 + 1 \).

This tells us several things about our parabola:

• The vertex is located at \(\left(\frac{1}{2}, 1\right)\), which is the peak point around which our parabola is symmetric.
• Since \(a = \frac{1}{2} > 0\), the parabola opens upwards.
• The line of symmetry is given by \(x = \frac{1}{2}\).

Understanding the qualities of a parabola helps in sketching its graph accurately. The vertex is a crucial part since it defines the height and center of the parabola's path.

Vertices and Foci

Vertices and foci are essential features of conic sections. For a parabola:

• The Vertex is the turning point of the curve. In our equation, the vertex is \(\left(\frac{1}{2}, 1\right)\). This vertex serves as the middle point that dictates the symmetry around which the parabola is shaped.
• The Focus provides information about the parabola's distinct curvature. Using the formula for the focus of a parabola in vertex form: \(y = a(x-h)^2 + k\),

The focus is positioned at \(\left(\frac{1}{2}, 1 + \frac{1}{4a}\right)\). Given that \(a = \frac{1}{2}\), the focus is found at \(\left(\frac{1}{2}, \frac{3}{2}\right)\).Finally, the directrix is a line at the opposite side of the focus, represented as:

• \(y = k - \frac{1}{4a}\).

This line in our case is \(y = \frac{1}{2}\).Each of these components helps construct a more precise structure of the parabola, ensuring a proper understanding of its position and shape.