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Chapter 10: Problem 65

Find the vertex, focus, and directrix of the parabola and sketch its graph. Use a graphing utility to verify your graph. $$x^{2}+6 y=0$$

Short Answer

Expert verified
The vertex is at (0,0), the focus is at (0,-1.5) and the directrix is y=1.5

Step by step solution

01

Convert to Standard Form

Rewrite the equation so that it resembles the standard form of the equation of a parabola: \(x^{2} = -6y\)
02

Identify and Calculate 'p'

From this form, we can see that -6 is -4p. Solving for 'p', we find that \(p = -6/(-4) = 1.5\).
03

Find the Vertex

Given that the formula is \(x^2 = -4py\), our vertex is at the origin, (0,0), because there is no horizontal or vertical shift.
04

Find the Focus

The focus of a parabola \(x^{2}=-4py\) is at (0, -p). Thus our focus is at (0, -1.5).
05

Find the Directrix

The directrix of the given parabolic equation is at y=p. Thus, our directrix is y = 1.5.
06

Sketch the Graph

Plot the vertex at the origin. Then plot the focus at (0, -1.5). Draw the line y=1.5 for the directrix. Sketch the parabola which opens downwards, with the vertex at the origin, and passing through the focus. Verify your graph using a graphical utility.

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Key Concepts

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

Vertex
In the world of parabolas, the vertex is a crucial point. Think of it as the point where the parabola either reaches its highest or lowest point. In a mathematical sense, it鈥檚 the tip of the U-shape. For our given equation, \(x^2 + 6y = 0\), after rearranging to \(x^2 = -6y\), we notice that it aligns with the standard form \(x^2 = -4py\).
  • Since our equation lacks any additional terms that would indicate a shift, this parabola is centered at the origin, meaning the vertex is at \((0, 0)\).
  • This is helpful because the vertex acts as a reference point for finding both the focus and the directrix, which we鈥檒l discuss in a bit.
  • The vertex is always located halfway between the focus and the directrix. It鈥檚 the point from which the parabola symmetrically opens.
Understanding the vertex location helps in sketching the graph. Always remember that the vertex plays a pivotal role in determining how the parabola opens and its overall orientation on the graph.
Focus
The focus of a parabola is a special point. It helps define the curve of the parabola. In essence, it is one of the focal points that results in the perfect shape of the parabola.

For a parabola described by \(x^2 = -4py\), the focus is situated at \((0, -p)\). From our previous calculations, \(p = 1.5\). Thus, the focus of our parabola is located at \((0, -1.5)\).
  • Imagine the parabola as a satellite dish, and the focus as the receiver located right above it.
  • Light or signals coming from the parabola will reflect off its curve and consistently meet at the focus point.
  • The focus doesn鈥檛 lie on the parabola but is a guiding point for its curve.
The role of the focus is to define how "wide" or "narrow" the parabola appears. The closer it is to the vertex, the narrower the parabola, and vice versa.
Directrix
The directrix is a straight line that provides balance along with the focus. It serves as a baseline against which the parabola is oriented. For our specific equation \(x^2 = -6y\), and knowing \(p = 1.5\), the directrix is at the line \(y = p\), thus \(y = 1.5\).
  • While the focus is a definite point beneath the vertex, the directrix lies above it for this parabola.
  • It does not graph on the parabola itself, but rather provides the perpendicular reference line.
  • Think of the directrix as a partner to the focus, creating a structure that ensures the parabola鈥檚 symmetry.
The directrix and focus work together. Any point on the parabola is equidistant to both the focus and the directrix. This foundational property helps in understanding how a parabola maintains its unique curved shape. Together, focus and directrix determine the geometry and position of the parabola in a plane.

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Most popular questions from this chapter

Identify the type of conic represented by the polar equation and analyze its graph. Then use a graphing utility to graph the polar equation. $$r=\frac{14}{14+17 \sin \theta}$$

Convert the polar equation to rectangular form. $$r=\frac{6}{2-3 \sin \theta}$$

Identify the type of conic represented by the equation. Use a graphing utility to confirm your result. $$r=\frac{6}{2+\sin \theta}$$

Find a polar equation of the conic with its focus at the pole. $$\begin{array}{cc} \text{Conic} & \text{Vertex or Vertices} \\\ \text{Parabola} &\left(1,-\frac{\pi}{2}\right)\end{array}$$

Convert the rectangular equation to polar form. Assume \(a<0\) $$\left(x^{2}+y^{2}\right)^{2}=9\left(x^{2}-y^{2}\right)$$
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