Problem 27 - College Algebra

Question

For the following exercises, rewrite the given equation in standard form, and then determine the vertex \((V),\) focus \((F),\) and directrix \((d)\) of the parabola. $$ x^{2}-4 x+2 y-6=0 $$

Step-by-Step Solution

Answer

Vertex: (2, 5), Focus: (2, 4.5), Directrix: y = 5.5

Step 1: Move Constant Term

To rewrite the equation in a form that reveals the parabola's properties, first move the constant term to the other side. The equation becomes \(x^2 - 4x = -2y + 6\).

Step 2: Complete the Square

To complete the square for the \(x\)-terms, 1. Take the coefficient of \(x\), which is \(-4\).2. Divide by 2 to get \(-2\).3. Square this to get \(4\).Add and subtract this square inside the equation: \(x^2 - 4x + 4 - 4 = -2y + 6\). This transforms to \((x-2)^2 - 4 = -2y + 6\).

Step 3: Simplify and Rearrange

Add 4 to both sides to simplify, yielding \((x-2)^2 = -2y + 10\). Then arrange to \((x-2)^2 = -2(y-5)\) to match the standard form of a parabola \((x-h)^2 = 4p(y-k)\). Here, \(h=2\), \(k=5\), so the vertex \(V(h, k) = (2, 5)\).

Step 4: Interpret Standard Form

In the equation \((x-2)^2 = -2(y-5)\), compare with the form \((x-h)^2 = 4p(y-k)\) to identify \(4p = -2\). Thus, \(p = -\frac{1}{2}\).

Step 5: Find Focus and Directrix

Given the vertex \(V(2, 5)\) and \(p = -\frac{1}{2}\), the focus \(F\) is \((2, 5 + p)\) or \(\left(2, 5 - \frac{1}{2}\right) = (2, 4.5)\). The directrix is \(y = 5 - p\) or \(y = 5 + \frac{1}{2} = 5.5\).

Key Concepts

Vertex Focus Directrix

Understanding the vertex, focus, and directrix of a parabola is crucial in mastering how to describe its orientation and dimension on a graph. The vertex represents the peak or the lowest point on a parabola, depending on its orientation:

• The vertex \( V(h, k) \) is a point where the parabola changes direction.
• The focus \( F \) is a point from which distances to any point on the parabola reflect the parabola’s shape.
• The directrix is a line perpendicular to the axis of symmetry that helps in defining the parabola’s geometry.

Each element provides a unique perspective on the parabola’s placement and its concave opening. In this example, with the vertex \( V(2, 5) \), focus \( F(2, 4.5) \), and directrix \( y = 5.5 \), these attributes describe a parabola opening downwards because \( p \) (the distance from the vertex to the focus) is negative.

Completing the Square

Completing the square is a technique used to transform a quadratic equation into its standard form. This reveals the properties of a parabola more clearly. The process involves:

• Identifying the coefficient of the \( x \) term, dividing it by 2, and squaring the result.
• Adding and subtracting this square within the equation to form a perfect square trinomial.

In our example, the equation \( x^2 - 4x + 4 - 4 = -2y + 6 \) simplifies to \( (x-2)^2 = -2(y-5) \). Completing the square allows us to clearly identify the transformation needed to find the vertex form of the equation.

Parabola Properties

Parabolas have distinctive properties that make them unique among conic sections. Understanding these properties can help easily predict the behavior of the graph. Key properties include:

• The axis of symmetry: This is a vertical line that passes through the vertex, meaning in our case, it is \( x = 2 \).
• The direction of opening: Based on our standard form transformation, \( y = 5.5 \) indicates the parabola opens downward due to the negative \( p \).
• Reflective property: The shape of the parabola ensures that rays coming parallel to the axis of symmetry will be reflected to pass through the focus.

These properties are essential when sketching or analyzing the curve without the need for plotting multiple points.

Parabola Equation Transformation

The transformation of a basic quadratic equation into a standard form parabola equation involves a few key steps that open up insights into the graph's geometry. The transition from \( x^2 - 4x + 2y - 6 = 0 \) to \( (x-2)^2 = -2(y-5) \) shows the structural changes:

• Shifts in position: The new vertex dictates the shift of the entire parabola from the origin.
• Orientation: Changes in the equation's coefficients reveal the opening's direction and width.

This transformation is a comprehensive way to visualize the parabola's path and how it situates itself in the Cartesian plane. Understanding how to manipulate equations into standard forms expands the methodology used in solving complex problems while harnessing the full descriptive power of their mathematical expressions.