Problem 31
Question
For the following exercises, graph the parabola, labeling the focus and the directrix. $$ x=\frac{1}{8} y^{2} $$
Step-by-Step Solution
Verified Answer
The parabola opens to the right with vertex at (0,0), focus at (2,0), and directrix x = -2.
1Step 1: Identify the conic section equation
The equation given is \(x = \frac{1}{8} y^2\). It resembles the standard form of a parabola equation \(x = \frac{1}{4p} y^2\), where the parabola opens along the x-axis.
2Step 2: Determine the value of p
In the general equation \(x = \frac{1}{4p} y^2\), we compare it with \(x = \frac{1}{8} y^2\) to identify \(\frac{1}{4p} = \frac{1}{8}\). Solving for \(p\), we get \(4p = 8\), thus \(p = 2\).
3Step 3: Find the vertex of the parabola
The given equation \(x = \frac{1}{8} y^2\) is centered at the origin, so the vertex of the parabola is at the point \((0, 0)\).
4Step 4: Determine the focus of the parabola
The focus of a parabola \(x = \frac{1}{4p} y^2\) is \((p, 0)\). Here, \(p = 2\), so the focus is at \((2, 0)\).
5Step 5: Determine the directrix of the parabola
The directrix of this parabola is a vertical line given by \(x = -p\). Since \(p = 2\), the directrix is the line \(x = -2\).
6Step 6: Sketch the graph
Draw the y-axis and x-axis. Plot the vertex at \((0, 0)\), the focus at \((2, 0)\), and the directrix as the line \(x = -2\). Sketch the parabola opening to the right, with its symmetry axis being the y-axis.
Key Concepts
Conic SectionsFocusDirectrixVertex
Conic Sections
Conic sections are curves obtained by intersecting a plane with a cone. There are four primary types of conic sections: circles, ellipses, parabolas, and hyperbolas. The shape we get depends on the angle at which the plane cuts through the cone.
In our case, the exercise dealt with parabolas, one of the four conic sections. Parabolas are unique because they have specific defining features such as a vertex, focus, and directrix, which guide their shape.
In mathematical context, parabolas follow certain standard forms, like the equation given in the exercise: \( x = \frac{1}{8} y^2 \), which is representative of a parabola opening along the x-axis.
In our case, the exercise dealt with parabolas, one of the four conic sections. Parabolas are unique because they have specific defining features such as a vertex, focus, and directrix, which guide their shape.
In mathematical context, parabolas follow certain standard forms, like the equation given in the exercise: \( x = \frac{1}{8} y^2 \), which is representative of a parabola opening along the x-axis.
Focus
The focus is a critical part of understanding parabolas and conic sections in general. It is a distinct point inside a parabola from which distances are measured to define the curve. What makes parabolas fascinating is that all points on the parabola are equidistant from the focus and a line called the directrix.
In our example, after identifying the equation \(x = \frac{1}{8} y^2\), we find that \(p = 2\). This tells us that the focus is at the point \((2, 0)\). The focus is always inside the curve of the parabola, pointing towards the direction in which it opens.
Remember, for any parabola given by the form \(x = \frac{1}{4p} y^2\), the coordinates of the focus are \((p, 0)\). This pinpoints its position relative to the origin.
In our example, after identifying the equation \(x = \frac{1}{8} y^2\), we find that \(p = 2\). This tells us that the focus is at the point \((2, 0)\). The focus is always inside the curve of the parabola, pointing towards the direction in which it opens.
Remember, for any parabola given by the form \(x = \frac{1}{4p} y^2\), the coordinates of the focus are \((p, 0)\). This pinpoints its position relative to the origin.
Directrix
The directrix is equally important when studying parabolas and other conic sections. It is a straight line and serves as a constant reference point for the parabola. Together with the focus, the directrix helps form the geometric definition of a parabola: the locus of points equidistant from the focus and the directrix.
In our equation \(x = \frac{1}{8} y^2\), we identify the directrix as a vertical line \(x = -2\). This is derived from the formula \(x = -p\), and since \(p = 2\), we position the directrix at \(x = -2\).
While the focus lies on one side of the vertex, the directrix lies on the opposite side. They provide a symmetric balance that is crucial to the precise shape of the parabola.
In our equation \(x = \frac{1}{8} y^2\), we identify the directrix as a vertical line \(x = -2\). This is derived from the formula \(x = -p\), and since \(p = 2\), we position the directrix at \(x = -2\).
While the focus lies on one side of the vertex, the directrix lies on the opposite side. They provide a symmetric balance that is crucial to the precise shape of the parabola.
Vertex
The vertex of a parabola acts as a turning point or the point where the curve changes direction. It is the midpoint between the directrix and the focus, making it an essential landmark in graphing and solving parabolas.
In the given exercise, the equation \(x = \frac{1}{8} y^2\) identifies that the parabola is centered at the origin \,\((0, 0)\). Therefore, the vertex is simply at \,\((0, 0)\).
The vertex is often the gateway to understanding the orientation and position of a parabola in space. Whether the parabola faces up, down, left, or right, hinges on the vertex's relative position to the focus and directrix. Knowing the vertex allows for precision when sketching or analyzing the behavior of the parabola.
In the given exercise, the equation \(x = \frac{1}{8} y^2\) identifies that the parabola is centered at the origin \,\((0, 0)\). Therefore, the vertex is simply at \,\((0, 0)\).
The vertex is often the gateway to understanding the orientation and position of a parabola in space. Whether the parabola faces up, down, left, or right, hinges on the vertex's relative position to the focus and directrix. Knowing the vertex allows for precision when sketching or analyzing the behavior of the parabola.
Other exercises in this chapter
Problem 31
For the following exercises, graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices
View solution Problem 31
Graph the given conic section. If it is a parabola, label the vertex, focus, and directrix. If it is an ellipse, label the vertices and foci. If it is a hyperbo
View solution Problem 31
For the following exercises, find the foci for the given ellipses. $$ 10 x^{2}+y^{2}+200 x=0 $$
View solution Problem 31
For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. $$ \frac{x^{2}}{49}-\frac{y^{2}}{16}=1 $$
View solution