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Chapter 10: Problem 33

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$x^{2}+6 y=0$$

Short Answer

Expert verified
The vertex of the parabola is at (0,0), the focus is at (0, -3/2), and the directrix is the line \(y = 3/2\).

Step by step solution

01

Convert to Standard Form

Rewrite the equation \(x^{2}+6 y=0\) in the standard form of a parabola. We can do this by isolating the \(y\)-term to one side which results in: \[x^{2} = -6y\] or further simplified to: \[x^{2} = 4py\] Comparing above two equations, it is clear that \(4p = -6\) and hence \(p = -3/2\).
02

Find the Vertex

For a parabola in the form \(x^{2} = 4py\), the vertex is always at the origin, (0,0).
03

Find the Focus and Directrix

The focus lies 'p' units above the vertex on the y-axis and the directrix is a horizontal line 'p' units below the vertex. Since in our case, \(p= -3/2\), then the focus is at (0, -3/2) and the directrix is the line \(y = 3/2\).
04

Sketch the Parabola

The vertex is at the origin (0,0), the focus is below the vertex at (0, -3/2), and the directrix is a horizontal line above the vertex at \(y = 3/2\). Given that the parabola opens downwards (as \(p\) is negative), we can sketch it accordingly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vertex
A parabola is a unique curve where every point is equidistant from a specific point, called the focus, and a line called the directrix. A key component of a parabola is its vertex. The vertex is the point where the parabola changes direction and can be seen as the "tip" or "peak" of the curve.
For a simple upward or downward opening parabola in the standard form \(x^2 = 4py\):
  • The vertex is located at (0, 0).
  • The equation highlights that the x-term is square, meaning the curve is symmetrical about the y-axis.
In the given exercise, the vertex is at the origin because the parabola takes the form \(x^2 = 4py\) with symmetric properties around the origin. The vertex is significant because it provides a reference point for understanding the parabola's other parts, like the focus and directrix.
Focus
The focus of a parabola is a fixed point that, along with the directrix, helps in defining the shape and position of the parabola. Every point on the parabola is equidistant from the focus and the directrix.
  • The location of the focus directly impacts how "narrow" or "wide" the parabola appears.
  • For a parabola expressed as \(x^2 = 4py\), the focus is located at the point (0, p) when the vertex is at the origin.
In our exercise, since \(p = -\frac{3}{2}\), the focus of the parabola is at (0, -\frac{3}{2}). This means the focus is situated below the vertex, indicating the parabola opens downward. The focus helps in understanding the direction in which the parabola opens and is crucial for sketching its precise shape.
Directrix
The directrix is an essential component associated with parabolas, providing a line reference against which distances are measured to define the shape. The directrix is a straight line, and its relationship with the focus is what determines how the parabola forms.
  • It is a horizontal line that lies opposite to the direction in which the parabola opens, in relation to the vertex.
  • For our standard form \(x^2 = 4py\), the directrix is located at \(y = -p\) when the vertex is at the origin.
For the given parabola, with \(p = -\frac{3}{2}\), the directrix is calculated as \(y = \frac{3}{2}\). This implies the directrix is above the vertex of the parabola. By situating this line opposite the focus, the directrix helps mark the boundary that balances the equidistant property from any point on the parabola to both the focus and the directrix.
Standard Form
The standard form of a parabola is a way of expressing its equation that makes identifying the vertex, focus, and directrix straightforward. It simplifies the understanding and manipulation of the parabola's properties.
  • The standard form for a vertical parabola is \(x^2 = 4py\).
  • The value of \(p\) indicates the distance between the vertex and the focus, as well as the vertex and the directrix.
In the context of the given exercise, the equation \(x^2 + 6y = 0\) becomes \(x^2 = -6y\) when rearranged, which fits into the standard form \(x^2 = 4py\) with \(4p = -6\). Thus, \(p = -\frac{3}{2}\), influencing the parabola's shape and direction. The standard form is particularly useful because once an equation is in this format, identifying the critical components of the parabola becomes a straightforward task.

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Most popular questions from this chapter

The planets travel in elliptical orbits with the sun at one focus. Assume that the focus is at the pole, the major axis lies on the polar axis, and the length of the major axis is \(2 a\) (see figure). Show that the polar equation of the orbit is \(r=a\left(1-e^{2}\right) /(1-e \cos \theta),\) where \(e\) is the eccentricity.

Convert the polar equation to rectangular form. $$r=-5 \sin \theta$$

Convert the polar equation to rectangular form. $$r=\frac{5}{1-4 \cos \theta}$$

Convert the polar equation to rectangular form. Then sketch its graph. $$\theta=3 \pi / 4$$

Determine whether the statement is true or false. Justify your answer. If \(\theta_{1}=\theta_{2}+2 \pi n\) for some integer \(n,\) then \(\left(r, \theta_{1}\right)\) and \(\left(r, \theta_{2}\right)\) represent the same point in the polar coordinate system.
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