Problem 77
Question
Write an equation for each parabola with vertex at the origin. Focus \(\left(-\frac{1}{2}, 0\right)\)
Step-by-Step Solution
Verified Answer
The equation is \( y^2 = -2x \).
1Step 1: Understand the Standard Form
A parabola with a horizontal axis of symmetry and a vertex at the origin is described by the equation \( y^2 = 4px \), where \( p \) is the distance from the vertex to the focus.
2Step 2: Identify the Distance to the Focus
From the vertex (0,0) to the focus \((-\frac{1}{2}, 0)\), the distance \( p \) is \(-\frac{1}{2} \), implying the parabola opens to the left.
3Step 3: Substitute the Distance into the Equation
Substitute \( p = -\frac{1}{2} \) into the standard equation to get \( y^2 = 4(-\frac{1}{2})x \). This simplifies to \( y^2 = -2x \).
4Step 4: Double-Check the Equation Orientation
Since \( p = -\frac{1}{2} \) is negative, the parabola indeed opens to the left, which matches the equation \( y^2 = -2x \).
Key Concepts
VertexFocusConic SectionsDistance from Vertex to Focus
Vertex
In a parabola, the vertex is one of the most crucial elements. It represents the point where the parabola changes direction.
For parabolas centered at the origin (0,0), the vertex indicates whether the parabola opens up, down, left, or right.
In the standard parabola equation forms:
For parabolas centered at the origin (0,0), the vertex indicates whether the parabola opens up, down, left, or right.
In the standard parabola equation forms:
- Vertical opening: \( x^2 = 4py \)
- Horizontal opening: \( y^2 = 4px \)
Focus
The focus of a parabola is a critical point that defines its shape and size. It is not on the curve itself but rather a point such that distances from it to any point on the parabola are equal to distances from the point on the parabola to a directrix (a not explicitly visible line).
A parabola is defined as the set of all points equidistant from the focus and the directrix.
For our given problem:
A parabola is defined as the set of all points equidistant from the focus and the directrix.
For our given problem:
- The focus is \( \left(-\frac{1}{2}, 0\right) \).
- This specific placement of the focus means that the parabola will open in the horizontal direction, either left or right, contingent on the value of \( p \).
Conic Sections
Parabolas are one of the four types of conic sections, which also include ellipses, circles, and hyperbolas. Conic sections arise by cutting a right circular cone with a plane. Parabolas occur when the plane is parallel to the cone's side.
Key features that are used to categorize and define conic sections include:
Key features that are used to categorize and define conic sections include:
- The vertex - the point where the conic section is widest or narrowest.
- The focus - a point used in the definition and formation of the shape.
- The directrix - a line used to define the conic section along with the focus.
Distance from Vertex to Focus
The distance from the vertex to the focus, usually denoted as \( p \), directly impacts the equation and direction of a parabola.
This distance can be positive or negative, indicating which way the parabola opens.
In the problem you encountered:
This distance can be positive or negative, indicating which way the parabola opens.
In the problem you encountered:
- \( p = -\frac{1}{2} \)
- This negative value means that the parabola opens to the left, as seen in the equation \( y^2 = 4px \), which becomes \( y^2 = -2x \).
Other exercises in this chapter
Problem 77
Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} x^{2}-x y+y^{2} &=5 \\ 2 x^{2}+x y-y^{2} &=10 \end{aligned}$$
View solution Problem 77
Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. $$x^{2}-2 x-y^{2}+2 y=4$$
View solution Problem 78
Solve each nonlinear system of equations analytically for all real solutions. $$\begin{aligned} &x^{2}+3 x y+y^{2}=5\\\ &x^{2}-2 x y-y^{2}=-7 \end{aligned}$$
View solution Problem 78
Write the equation in standard form for a hyperbola centered at ( \(h, k\) ). Identify the center and vertices. $$y^{2}+4 y-x^{2}+2 x=6$$
View solution