Problem 13
Question
Find the standard equation of each parabola from the given information. Assume that the vertex is at the origin. Focus is \((-4,0)\)
Step-by-Step Solution
Verified Answer
The standard equation is \( y^2 = -16x \).
1Step 1: Understanding the Parabola Orientation
Since the focus is at (-4,0) and the vertex is at the origin (0,0), the parabola opens horizontally because the focus is located on the x-axis. Specifically, it opens to the left because the x-coordinate of the focus is negative.
2Step 2: Standard Form of a Parabola Equation
The standard form of a parabola that opens horizontally (with vertex at (0,0)) is given by \( y^2 = 4px \). Here, \( p \) is the distance from the vertex to the focus.
3Step 3: Calculating p
In this problem, the focus is (-4,0). Therefore, the distance \( p \) from the vertex (0,0) to the focus (-4,0) is 4 units. Since the parabola opens to the left, \( p = -4 \).
4Step 4: Substitute p into the Standard Equation
Substituting \( p = -4 \) into the equation \( y^2 = 4px \) gives \( y^2 = 4(-4)x \), so the equation becomes \( y^2 = -16x \).
Key Concepts
VertexFocusDistance from Vertex to FocusStandard FormHorizontal Orientation
Vertex
In the context of a parabola, the vertex is a crucial point on the graph. It is the point where the parabola reaches its maximum or minimum value. In our specific exercise, the vertex is positioned at the origin, denoted by the coordinates
(0,0). This location is foundational because it serves as the turning point of the parabola. - The vertex is always equidistant from the focus and the directrix, forming a perfect axis of symmetry for the parabola.
- When the vertex is at the origin, computations and equations become much simpler, because shifts or transformations like vertical or horizontal moves are not required.
Focus
The focus of a parabola is a point inside the curve that helps in defining the parabola's shape and orientation. It's where light or signals theoretically converge if the parabola were used as a reflector.
- For our exercise, the focus is located at the point (-4,0).
- The relation between the vertex and the focus determines the direction in which the parabola opens. A focus to the right or left of the vertex indicates a horizontal parabola.
- The position of the focus relative to the vertex also affects how narrow or wide the parabola appears.
Distance from Vertex to Focus
The distance from the vertex to the focus, denoted as **p**, is a key parameter in forming the equation of the parabola. This measurement describes how far the focus is from the vertex along the axis of symmetry.
- In our case, this distance is 4 units because the focus is
(-4,0). - Since the parabola opens to the left, **p** is negative, giving us
p = -4. The negative sign indicates the leftwards opening. - This distance directly influences the coefficient in the parabola's equation.
Standard Form
The standard form of a parabola's equation offers a streamlined approach to describing its shape and behavior mathematically. For a parabola opening horizontally, the standard form used is
y^2 = 4px. - Here,
4pis a factor that scales the parabola, connecting directly with the distance **p** from the vertex to the focus. - The standard form allows for quick identification of the parabola's direction (leftward or rightward) by examining the sign of
p. - If the vertex is at the origin, any computations become easier and do not involve shifting the graph.
Horizontal Orientation
In this parabola, the horizontal orientation is characterized by the parabola opening leftward due to the location of the focus.
- This orientation is determined by the focus's placement on the x-axis and to the left of the vertex.
- The equation
y^2 = -16xconfirms the leftward opening, as signified by the negative4pvalue in our expression. - A horizontally oriented parabola intersects the y-axis and has reflective properties differing from vertical parabolas.
Other exercises in this chapter
Problem 13
Name the conic or limiting form represented by the given equation. Usually you will need to use the process of completing the square. $$ 4 x^{2}-24 x+36=0 $$
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Sketch the limaçon \(r=2-3 \cos \theta\), and find the area of the region inside its large loop.
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In each of Problems 11-16, sketch the graph of the given Cartesian equation, and then find the polar equation for it. \(x-y=0\)
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In each of Problems 1-20, a parametric representation of a curve is given. (a) Graph the curve. (b) Is the curve closed? Is it simple? (c) Obtain the Cartesian
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