/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Determine the equation in standa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 54

Determine the equation in standard form of the parabola that satisfies the given conditions. Focus at (0,-5)\(;\) directrix \(y=5\)

Short Answer

Expert verified
The equation of the parabola is \(y = -0.05x^2\).

Step by step solution

01

Find the vertex

The vertex is midway between the focus and directrix, the y-coordinate of which can be calculated as: \(k = (y_{directrix} + y_{focus})/2 = (5 + (-5))/2 = 0\). The x-coordinate is the same as the x-coordinate of the focus, thus the vertex is: \(v = (h, k) = (0, 0)\).
02

Determine the orientation

The focus is below the directrix, so the parabola will open downwards.
03

Find the value of \(a\)

We find \(a\) from the relationship \(a = -1/(4f)\). The distance from the vertex to the focus or directrix is \(f = 5\), so \(a = -1/(4*5) = -0.05\).
04

Write out the equation

Now that we have the vertex (h, k) and \(a\), we can plug these values into the vertex form of a parabola equation: \(y = a(x-h)^2 + k\). Substituting gives the equation of the parabola as \(y = -0.05x^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Focus and Directrix
The focus and directrix are crucial components when describing a parabola. They're like the GPS directions that help you locate how and where a parabola sits on the coordinate plane.
  • The **focus** is a point inside the parabola that is used to define and construct the curve. It's like the heart of the parabola, where all the curves of the parabola seem to "focus" around.
  • The **directrix** is like an imaginary line that helps to steer the curve in its fixed path. It is always perpendicular to the axis of symmetry of the parabola.

The relationship between these two components is fundamental. For any point on the parabola, the distance to the focus is equal to the shortest distance to the directrix.
In our exercise, these distances from each point are equal, with the focus located at \((0, -5)\) and the directrix being the line \(y = 5\). Together, they guide this particular parabola to be centered perfectly between these two features.
Vertex of a Parabola
The vertex of a parabola is its peak or its lowest point depending on the orientation of the curve. It's like the throne of the parabola, the point where it majestically sits and determines the direction it opens towards.
In mathematical terms, the vertex can be found exactly midway between the focus and directrix. To find it, you average the coordinates of the focus and the value of the directrix.
  • The **x-coordinate** of the vertex is the same as the x-coordinate of the focus.
  • The **y-coordinate** is the midpoint between the y-coordinates of the focus and directrix.

For the given problem, this magical spot known as the vertex is located at \((0, 0)\), precisely in the center horizontally and vertically of the configurations given. This symmetry plays a key role in defining the shape and equation of the parabola.
Orientation of a Parabola
Understanding the orientation of a parabola is like knowing whether to draw a smile or a frown. This direction tells us how the arms of the parabola stretch out.
A parabola can open either upward or downward if it is vertical, or it can open right or left if it is horizontal.
In the case of vertical parabolas (like our example), the orientation depends on the relative positioning of the focus and directrix:
  • If the focus is below the directrix, as in our exercise, the parabola opens **downwards**.
  • Conversely, if the focus were above the directrix, it would open **upwards**.

This specific parabola opens downwards, which aligns with the calculated vertex and derived equation. Recognizing how a parabola opens is essential to correctly plotting the curve and verifying the construction of the parabola's equation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Identify and graph the conic section given by each of the equations. $$r=\frac{4}{2+5 \cos \theta}$$

Graph each equation using a graphing utility. $$x^{2}-x y+\frac{1}{4} y^{2}-2 y=0$$

Identify and graph the conic section given by each of the equations. $$r=\frac{16}{8-4 \cos \theta}$$

The eccentricity of an ellipse is defined as \(e=\frac{c}{a}\) \(\left(=\frac{\sqrt{a^{2}-b^{2}}}{a}\right),\) where \(a, b,\) and \(c\) are as defined in this section. since \(0 < c < a,\) the value of \(e\) lies between 0 and 1 In ellipses that are long and thin, \(b\) is small compared to \(a,\) so the eccentricity is close to \(1 .\) In ellipses that are nearly circular, \(b\) is almost as large as \(a,\) so the eccentricity is close to \(0 .\) What is the eccentricity of the ellipse with equation \(\frac{x^{2}}{9}+\frac{y^{2}}{25}=1 ?\) Does this ellipse have a greater or lesser eccentricity than the ellipse with equation \(\frac{x^{2}}{16}+\frac{y^{2}}{25}=1 ?\)

In this set of exercises, you will use hyperbolas to study real-world problems. Astronomy The path of a certain comet is known to be hyperbolic, with the sun at one focus. Assume that a space station is located 13 million miles from the sun and at the center of the hyperbola, and that the comet is 5 million miles from the space station at its point of closest approach. Find the equation of the hyperbola if the coordinate system is set up so that the sun lies on the \(x\) -axis and the origin coincides with the center of the hyperbola.
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.