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Chapter 11: Problem 47

Find an equation for the parabola that has its vertex at the origin and satisfies the given condition(s). Directrix has \(y\) -intercept 6

Short Answer

Expert verified
The equation is \( x^2 = -24y \).

Step by step solution

01

Understand the Parabola's Properties

We need an equation for a parabola with a vertex at the origin and a directrix at a line with a y-intercept of 6. A parabola is defined as the set of points equidistant from the focus and the directrix. For this exercise, the vertex (0, 0) implies that the parabola opens upwards or downwards since its vertex is at the origin, depending on the position of the directrix.
02

Determine the Directrix Equation

The directrix is a horizontal line with a y-intercept at 6, so its equation is \( y = 6 \). Since it lies above the vertex at the origin (0,0), the parabola opens downwards.
03

Determine the Focus

The distance from the vertex to the directrix is half the value needed to find the focus. Since the directrix is \( y = 6 \), the distance to the vertex is 6. Thus, the focus, which is equidistant from the vertex, is at (0, -6) because the parabola opens downwards.
04

Write the Parabola Equation Form

For a parabola opening vertically, the standard equation is \( x^2 = 4py \), where \( p \) is the distance from the vertex to the focus. Here, \( p = -6 \) as the parabola opens downwards. This means the parabola's equation is \( x^2 = -24y \), because \( 4p = 4(-6) = -24 \).

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Key Concepts

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

Vertex Form
In the world of parabolas, the vertex form is incredibly useful for identifying key features of the parabola, such as the vertex itself. The vertex form of a parabola's equation is written as \( y = a(x-h)^2 + k\), where \( (h, k) \) represents the vertex.
  • Here, \( h \) and \( k \) are the x and y coordinates of the vertex.
  • For example, a vertex form of \( y = 2(x - 3)^2 + 4 \) indicates a vertex at \( (3, 4) \)
For our specific exercise, the vertex form can be simplified further since the vertex is at the origin, turning into \( y = ax^2 \). This makes solving easier and provides a direct path to forming the parabola equation when additional conditions like the directrix are considered.
Directrix
The directrix is a fixed line used in defining a parabola. It's a geometric construct that helps in understanding how a parabola is shaped. A parabola can be seen as the locus of points that are equidistant from a fixed point called the **focus** and a line known as the **directrix**.
In our example, the directrix is the line identified by the equation\( y = 6\). It lies above the vertex at \( (0, 0) \), indicating that our parabola opens downwards.
  • The distance from the vertex to this line helps determine the focus of the parabola
  • For every point on the parabola, this distance to the directrix is equal to the distance to the focus.
This equidistant property plays a critical role in the symmetry and shape of parabolas.
Focus of a Parabola
The focus of a parabola is key to its geometric definition, helping shape its form. This fixed point is as crucial as the directrix in the parabolic structure.
For our exercise, the focus, along with the directrix \( y = 6 \), determines the orientation and direction of the parabola. With the vertex at the origin \( (0, 0) \) and a directrix at \( y = 6 \), the parabola opens downwards, positioning the focus at \( (0, -6) \).
  • This position is because the parabola must be equidistant from the focus and the directrix.
  • The distance between the vertex and the focus (\( p \)) is 6 units, calculated by taking half of the distance from the vertex to the directrix.
Understanding the role of the focus helps in deriving the standard form of the parabola's equation.
Conic Sections
Conic sections are curves obtained by slicing a cone at various angles. These include parabolas, ellipses, circles, and hyperbolas. Each type differs by the nature and angle of intersection.
Parabolas specifically arise when the cone is cut parallel to its side, leading to their distinctive U-shaped curve. Here are some key characteristics:
  • Symmetry around a central axis, which is aligned with their vertex.
  • A unique relationship between every point on the curve with its focus and directrix.
Understanding parabolas as conic sections allows one to explore their deeper geometrical properties and applications. From gravitational paths to satellite dishes, the utility of parabolas spans both practical and theoretical domains.

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Most popular questions from this chapter

An equation of a parabola is given. (a) Find the vertex, focus, and directrix of the parabola. (b) Sketch a graph showing the parabola and its directrix. $$(y+5)^{2}=-6 x+12$$

The planets move around the sun in elliptical orbits with the sun at one focus. The point in the orbit at which the planet is closest to the sun is called perihelion, and the point at which it is farthest is called aphelion. These points are the vertices of the orbit. The earth's distance from the sun is \(147,000,000 \mathrm{km}\) at perihelion and \(153,000,000 \mathrm{km}\) at aphelion. Find an equation for the earth's orbit. (Place the origin at the center of the orbit with the sun on the \(x\) -axis.) (IMAGE CAN'T COPY)

The graph of the equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) with \(a>b>0\) is an ellipse with vertices (________ , ________ )and (_______ ,________ ) and foci \((\pm c, 0),\) where \(c=\) ________ . So the graph of \(\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1\) is an ellipse with vertices (_______ , _______ ) and (_______,_______) and foci (_______ , _______ ) and (_______,_______).

The polar equation of an ellipse can be expressed in terms of its eccentricity \(e\) and the length \(a\) of its major axis. (a) Show that the polar equation of an ellipse with directrix \(x=-d\) can be written in the form $$r=\frac{a\left(1-e^{2}\right)}{1-e \cos \theta}$$ [Hint: Use the relation \(a^{2}=e^{2} d^{2} /\left(1-e^{2}\right)^{2}\) given in the proof on page 825 .] (b) Find an approximate polar equation for the elliptical orbit of the earth around the sun (at one focus) given that the eccentricity is about 0.017 and the length of the major axis is about \(2.99 \times 10^{8} \mathrm{km}\).

An equation of a hyperbola is given. (a) Find the vertices, foci, and asymptotes of the hyperbola. (b) Determine the length of the transverse axis. (c) Sketch a graph of the hyperbola. $$\frac{y^{2}}{9}-\frac{x^{2}}{16}=1$$
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