Problem 30 - Algebra 2

Question

Write an equation for each parabola described below. Then draw the graph. focus \((-4,-2),\) directrix \(x=-8\)

Step-by-Step Solution

Answer

The equation is (y+2)^2 = 8(x+6) . Draw the parabola with vertex (-6, -2) and opening to the right.

Step 1: Understand the Problem

We are given a focus at (-4,-2) and a directrix at x = -8 . We need to find the equation of the parabola that satisfies these conditions.

Step 2: Use the Definition of Parabola

A parabola is the set of all points that are equidistant from the focus and the directrix. For our parabola, any point (x, y) on the parabola will have the same distance to the focus (-4, -2) and the line x = -8 (the directrix).

Step 3: Apply the Distance Formula

The distance from a point (x, y) to the focus (-4, -2) is \(\sqrt{(x + 4)^2 + (y + 2)^2}\). The distance from (x, y) to the directrix x = -8 is \( |x + 8| \). Set these two distances equal:\[\sqrt{(x + 4)^2 + (y + 2)^2} = |x + 8| \]

Step 4: Square Both Sides

To eliminate the square root, square both sides of the equation:\[(x + 4)^2 + (y + 2)^2 = (x + 8)^2\]

Step 5: Expand Both Sides

Expand both sides of the equation:\[(x^2 + 8x + 16) + (y^2 + 4y + 4) = x^2 + 16x + 64\]

Step 6: Simplify and Solve for the Parabola Equation

Cancel x^2 from both sides and combine like terms:\[8x + 16 + y^2 + 4y + 4 = 16x + 64\]Rearrange terms:\[y^2 + 4y + 12 = 8x + 48\]This equation can be simplified further.

Step 7: Express in Standard Parabolic Form

Rewriting the final equation to isolate x:\[y^2 + 4y + 12 - 48 = 8x\]Cancelling out and rearranging gives:\[y^2 + 4y - 36 = 8x\]This is the equation of the parabola.

Step 8: Graph the Parabola

The equation y^2 + 4y = 8x + 36 represents a parabola with a horizontal orientation because we can rearrange it to (y + 2)^2 = 8(x + 6). The vertex form indicates this parabola opens to the right from its vertex. Plotting the vertex and the orientation of the curve gives the graph.

Key Concepts

Focus and Directrix

In geometry, a parabola is a set of points equidistant from a fixed point called the "focus" and a line called the "directrix." This property is the foundation of many operations related to parabolas. For the problem at hand, the focus is given as (-4, -2) and the directrix as x = -8.

The significance of the focus and directrix is that they help define the parabola's shape and position. By definition, every point on the parabola is at an equal distance from both the focus and the directrix. This equilibrium is what gives a parabola its distinctive curved shape, and understanding this is fundamental for working with parabolas.

• The focus is located inside the parabola, guiding the shape's curvature.
• The directrix is a fixed line outside the parabola and helps keep the curve oriented correctly.

Knowing these elements is key to identifying and graphing the equation of the parabola. This relationship guides us into deriving the specific equation that represents our parabola.

Equation of a Parabola

To find the equation of a parabola given a focus and a directrix, we use their special relationship with points on the parabola. The general equation of a parabola is rooted in setting the distance from any point on the curve to the focus equal to its distance to the directrix.

For a parabola with a horizontal orientation like in our problem, we typically express the equation in terms of x.

The standard form for a horizontally oriented parabola is (y - k)^2 = 4p(x - h), where (h, k) is the vertex of the parabola, and p is the distance from the vertex to the focus (or the negative of that to the directrix). Our derivation ultimately boils down to turning the symbolic relationship into a valid algebraic expression by:

• Identifying key points and distances from the focus and directrix.
• Applying the quadratic formula structure.
• Simplifying via algebraic manipulations to find the final parabolic equation.

Graphing Parabolas

Graphing a parabola effectively requires understanding its orientation, vertex, and the role of the focus and directrix. Once the equation (y + 2)^2 = 8(x + 6) has been derived from the provided information, it gives a definitive picture of the parabola's shape.

Key steps to graphing:

• Determine the Vertex: The vertex can be identified by transforming the parabola equation into its vertex form, revealing that here the vertex is at (-6, -2).
• Identify the Direction: Because the equation contains (y + 2)^2, the parabola opens horizontally to the right, consistent with the focus being to the right of the directrix.
• Sketch the Focus and Directrix: The focus and directrix positioning supports determining the parabola's path. This results in a visualization of how every point is equidistant to the focus and directrix.

By following these steps, the graph becomes a visual representation of the mathematical relationships inherent in the given problem.

Distance Formula in Geometry

The distance formula is key to applying the parabolic properties that utilize focus and directrix. It allows one to calculate the precise distance between two points in a plane, fundamental for nearly all geometry problems dealing with curves and lines.

The standard distance formula is \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
For our parabola, the focus (-4, -2) is used against the general points (x, y), calculating the distance from the parabola's focus. Furthermore, for the perpendicular distance from a point to a vertical line directrix (say x = -8), it becomes simply \( |x + 8| \).Understanding this formula boosts problem-solving abilities by:

• Confirming the definition of a parabola using equidistant properties.
• Allowing algebraic equivalences that are essential in deriving the parabola's equation.

Mastering this basic yet pivotal tool is critical for students tackling not just parabolas, but a wide array of geometric challenges.