Problem 19 - Precalculus: Mathematics for Calculus

Question

An equation of a parabola is given. (a) Find the vertex, focus, and directrix of the parabola. (b) Sketch a graph showing the parabola and its directrix. $$y^{2}-6 y-12 x+33=0$$

Step-by-Step Solution

Answer

Vertex at (2, 3), focus at (5, 3), directrix at x = -1; parabola opens right.

Step 1: Rewrite the Equation

First, we need to manipulate the given equation to identify the standard form of a parabola. The equation given is:\[ y^{2} - 6y - 12x + 33 = 0 \]We should move the terms around to complete the square for the terms involving \( y \).

Step 2: Complete the Square

Focus on completing the square for the \( y \) terms. Start with the terms \( y^{2} - 6y \):Add and subtract \( 9 \) (which is \((\frac{-6}{2})^2\)):\[ y^{2} - 6y + 9 - 9 = (y - 3)^{2} - 9 \]Substitute back into the equation:\[ (y - 3)^2 - 9 - 12x + 33 = 0 \]

Step 3: Simplify the Equation

Simplify the equation by moving all constant terms to the right side:\[ (y - 3)^{2} = 12x - 24 \]Then simplify further:\[ (y - 3)^{2} = 12(x - 2) \]

Step 4: Identify the Vertex, Focus, and Directrix

The equation \((y - 3)^{2} = 12(x - 2)\) represents a parabola that opens to the right. Its standard form is \((y - k)^{2} = 4p(x - h)\), where \((h, k)\) is the vertex, and \(p\) is the distance from the vertex to the focus or directrix.- Vertex: \((2, 3)\)- From \(4p = 12\), solve for \(p\):\[ p = \frac{12}{4} = 3 \]- Focus: \((h + p, k) = (2 + 3, 3) = (5, 3)\)- Directrix: \(x = h - p = 2 - 3 = -1\)

Step 5: Sketch the Graph

Sketch the graph of the parabola with vertex at \((2, 3)\), focus at \((5, 3)\), and directrix at \(x = -1\). The parabola opens to the right because the \(x\) term is isolated. Indicate the direction of the parabola and plot the vertex, focus, and directrix accurately.

Key Concepts

Vertex

In a parabola, the vertex is a crucial point that can tell us a lot about its shape and location. The vertex is either the highest or lowest point of the parabola, depending on its orientation. For the parabola given in the exercise, the vertex is found from the equation \[(y - 3)^{2} = 12(x - 2)\]
This form makes it easy to spot the vertex at \((h, k)\), which is \((2, 3)\). This means the vertex is located 2 units to the right along the x-axis and 3 units up along the y-axis.

• The vertex acts as a hinge or turning point for the parabola.
• It is the point where the parabola changes direction.
• For horizontal parabolas, the vertex shows the leftmost or rightmost point.

Understanding the vertex can help you quickly sketch and analyze parabolas.

Focus

The focus of a parabola is another important point that lies along the axis of symmetry, a line that runs through the vertex. It helps define the parabola shape and reflects how the parabola converges or diverges. For the equation given, once simplified to \((y - 3)^{2} = 12(x - 2)\),
we can calculate the focus. The focus for this parabola is at \((h + p, k)\):

• From \(4p = 12\), solve \(p = 3\).
• Adding \(p\) to \(h\): \(2 + 3 = 5\), so the focus is at \((5, 3)\).

The focus lies on the interior of the parabola and is used to define the direct path that causes lines parallel to the axis of symmetry to focus at this point. Understanding the focus helps when sketching the parabola and predicting how it behaves.

Directrix

The directrix of a parabola is a straight line that lies outside the parabola. It helps with understanding the geometric properties of this curve. It is always perpendicular to the axis of symmetry and has an important relationship with the focus.

• For the equation\((y - 3)^{2} = 12(x - 2)\), the directrix is found by subtracting \(p\) from \(h\):\(x = h - p\).
• This is calculated as \(x = 2 - 3 = -1\).

Here, the directrix is the vertical line \(x = -1\). This line provides additional reference for constructing the parabola, because the distance to a point on the parabola from the directrix is equal to the distance to the focus.

Completing the Square

Completing the square is a powerful algebraic method used to transform quadratic equations into a form that reveals important properties of parabolas. Let's break down the process:

• Start with the quadratic terms in \(y\): \(y^{2} - 6y\).
• Take half of the coefficient of \(y\), which is \(-6\), and square it: \((\frac{-6}{2})^2 = 9\).
• Add and subtract \(9\) inside the equation to complete the square:\(y^{2} - 6y + 9 = (y - 3)^{2} - 9\).

This technique simplifies the equation, allowing easy identification of the parabola's vertex and transforming the complex polynomial into a manageable form. Completing the square is essential for turning general quadratic equations into a standard form, making it much easier to graph and analyze the parabolic plot.