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Chapter 12: Problem 6

Conic Construction Problem 2: Plot on graph paper the conic with focus \((0,0),\) directrix \(x=-6,\) and eccentricity \(e=1 .\) Plot points for which the distance \(d_{1}\) from the directrix equals \(3,6,10,\) and \(20 .\) Connect the points with a smooth curve. Which conic section have you graphed?

Short Answer

Expert verified
Points are plotted that are equidistant from the focus (0,0) and the directrix (x=-6), forming a parabola with the vertex at (0,0).

Step by step solution

01

Understanding the Conic Definition

Let's start by understanding the conic section definition given. The set of points for a conic section is defined by a fixed point called the focus (in this case, (0,0)), a line called the directrix (in this case, x=-6), and the eccentricity (e). For a distance d1 from any point on the conic to the directrix, the distance to the focus will be e * d1. Since the eccentricity (e=1) for this conic, the distance from any point on the conic to the focus is the same as the distance from the point to the directrix.
02

Plotting Points

To plot points on the graph, we need to find points such that the distance to the focus (0,0) is equal to the distance to the directrix (x=-6). For the given distances d1=3, 6, 10, and 20, the points on the conic will be at the same horizontal distances from the directrix; since x=-6, these points will be located at x=-3, 0, 4, and 14, respectively. Their y-coordinates will be 0, since these particular points lie on the horizontal line passing through the focus.
03

Plotting Additional Points

Apart from horizontal points, we must find other points that satisfy the distance condition. For a given d1, we use Pythagorean theorem to find other points. For example, if d1=3, the points will form a circle with radius 3 centered at the focus (0,0). Similarly, for d1=6, the circle will have radius 6, and so on. We can plot several such points for each value of d1 to get an accurate shape of the conic.
04

Connecting the Points

After plotting points for each of the given distances and some in between, join those points with a smooth curve to form the conic section.
05

Identifying the Conic Section

The conic section with a focus at the origin, a vertical directrix to the left of the focus, and an eccentricity of 1 is a parabola. The points that are at equal distances from the focus and directrix lie on the parabola, thus the conic section graphed is a parabola.

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Key Concepts

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

Parabola
A parabola is one of the fundamental shapes under the umbrella of conic sections. It's unique in that it's defined by a single focus point and a line called the directrix. Imagine the focus as a broadcasting antenna sending signals in all directions, and consider the directrix as a boundary wall.

The parabola's special property is that any signal from the focus reflects off the curve and travels the same distance again to the directrix, which translates into a mathematical property: any point on the parabola is equidistant to the focus and the directrix. Now, this is precisely what was exploited in the given exercise. By plotting the points that are the same distance from the focus at (0,0) and the directrix at x=-6, one would draw a perfect outline of a parabola. When these distances are plotted using a smooth curve, the characteristic 'U' shape becomes apparent.
Eccentricity of Conic Sections
The term 'eccentricity' may sound like it's about how strange something is, but in geometry, it's a way of quantifying the 'roundness' of a conic section. Essentially, eccentricity is a measure of how much a conic deviates from being a perfect circle. For all conic sections, this value is denoted by 'e'.

For a circle, 'e' is zero, because there is no deviation鈥攊t鈥檚 the same distance from center to edge all the way around. As this value increases, the shape stretches out: an ellipse has an eccentricity between 0 and 1, a parabola has an eccentricity exactly equal to 1, and a hyperbola has an eccentricity greater than 1. In the exercise, an eccentricity of 1 points us directly to the conclusion that the conic section we're plotting is a parabola.
Directrix and Focus of Conics
Think of the focus and the directrix of conic sections as the architects of their shapes. The focus serves as a reference point, while the directrix acts as a guiding line against which distances are measured. These two components work together to mold the conic section into its definitive shape.

In the context of the original problem, the focus at (0,0) and the directrix at x=-6 establish a kind of symmetry axis for the parabola. As you learned from the step by step solution, the parabola is the set of all points that maintain an equal distance to the focus and the directrix, given the eccentricity鈥攚hich in this case is 1. This simple rule dictates the parabola's elegant curve and ensures every point on it abides by the reflective property central to its definition.

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Most popular questions from this chapter

For Problems \(11-20\) a. Identify the conic section. b. Calculate four radii and the eccentricity. c. Plot the graph. Sketch the result. $$\frac{x^{2}}{289}+\frac{y^{2}}{64}=1$$

Mars Orbit Problem: Mars is in an elliptical orbit around the Sun, with the Sun at one focus. The aphelion (the point farthest from the Sun) and the perihelion (the point closest to the Sun) are 155 million miles and 128 million miles, respectively, as shown in 1\. 41 (not to scale) a. How long is the major axis of the ellipse? What is the major radius? b. Find the focal radius and the minor radius of the ellipse. c. Write a Cartesian equation for the ellipse, with the center at the origin and the major axis along the \(x\) -axis. d. At the two equinoxes (times of equal day and night), the angle at the Sun between the major axis and Mars is \(90^{\circ} .\) At these times, what is the value of \(x ?\) How far is Mars from the Sun? e. Find the eccentricity of the ellipse. f. How far from the Sun is the closer directrix of the ellipse? g. Write parametric equations for the ellipse. Plot the graph using parametric mode. Zoom appropriately to make equal scales on the two axes. h. The ellipse you plotted in part g looks almost circular. How do the major and minor radii confirm this? How does the eccentricity confirm this?

a. Name the conic section simply by looking at the Cartesian equation. b. Sketch the graph. c. Transform the given equation to an equation of the form $$ A x^{2}+B x y+C y^{2}+D x+E y+F=0 $$ d. Plot the Cartesian equation using the result of part c. Does it agree with part b? $$\left(\frac{x-4}{10}\right)^{2}+\left(\frac{y-2}{10}\right)^{2}=1$$

a. Name the conic section simply by looking at the Cartesian equation. b. Sketch the graph. c. Transform the given equation to an equation of the form $$ A x^{2}+B x y+C y^{2}+D x+E y+F=0 $$ d. Plot the Cartesian equation using the result of part c. Does it agree with part b? $$(x+6)-1.5(y-3)^{2}=0$$

a. Name the conic section simply by looking at the Cartesian equation. b. Sketch the graph. c. Transform the given equation to an equation of the form $$ A x^{2}+B x y+C y^{2}+D x+E y+F=0 $$ d. Plot the Cartesian equation using the result of part c. Does it agree with part b? $$\left(\frac{x-3}{2}\right)^{2}+\left(\frac{y-1}{4}\right)^{2}=1$$
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