/*! 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 5 Find the vertex, focus, and dire... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 5

Find the vertex, focus, and directrix of the parabola, and sketch the graph. $$ (x-3)^{2}=8(y+1) $$

Short Answer

Expert verified
Vertex: (3, -1); Focus: (3, 1); Directrix: y = -3.

Step by step solution

01

Identify the Parabola's Standard Form

The given parabola equation \((x-3)^2 = 8(y+1)\) is in the form \((x-h)^2 = 4p(y-k)\), which represents a vertical parabola. Here, the vertex is \((h, k)\).
02

Determine the Vertex

From the equation \((x-3)^2 = 8(y+1)\), compare it with the standard form \((x-h)^2 = 4p(y-k)\). We have \(h = 3\) and \(k = -1\). Thus, the vertex of the parabola is \((3, -1)\).
03

Calculate the Value of p

In the standard form \((x-h)^2 = 4p(y-k)\), we equate \(4p = 8\). Solving for \(p\), we find \(p = 2\).
04

Find the Focus

The focus is located at \((h, k + p)\). With \(h = 3\), \(k = -1\), and \(p = 2\), the focus is \((3, 1)\).
05

Determine the Directrix

The directrix of a vertical parabola \((x-h)^2 = 4p(y-k)\) is given by \(y = k - p\). Plugging in the values, \(y = -1 - 2 = -3\), so the directrix is \(y = -3\).
06

Sketch the Graph

The vertex is at \((3, -1)\), the focus at \((3, 1)\), and the directrix at \(y = -3\). The parabola opens upwards (since \(p > 0\)). Plot the vertex and focus, draw the directrix as a horizontal line, and sketch the parabola opening upwards.

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.

Vertex
The vertex of a parabola is a key point where the parabola changes direction. In the equation \[(x-3)^2 = 8(y+1),\]we identify the vertex by comparing it to the standard form of a vertical parabola: \[(x-h)^2 = 4p(y-k).\]
  • Here, \(h = 3\) and \(k = -1\).
  • Thus, the vertex is located at \((3, -1)\).
The vertex represents either the highest or lowest point on the graph of a parabola. For this specific parabola, it is the lowest point given that it opens upwards. The coordinates of the vertex are crucial for sketching the parabola and understanding its orientation.
Focus
The focus of a parabola is a point inside the curve that helps in defining its shape and direction. From the equation \((x-3)^2 = 8(y+1)\), we've already determined that \(p=2\), which helps find the focus.
  • The focus for a vertical parabola can be found using the formula \((h, k+p)\).
  • Substituting the values: \(h = 3\), \(k = -1\), and \(p = 2\), we find the focus to be \((3, 1)\).
The focus is essential because it lies along the axis of symmetry of the parabola. Rays of light or other signals originating from the focus will reflect off the parabola and travel in specific directions, which is why this point is of great significance in applications such as satellite dishes and headlights.
Directrix
The directrix is a line perpendicular to the axis of symmetry of the parabola and plays an essential role in its geometric definition. For the parabola \[(x-3)^2 = 8(y+1),\]the equation of the directrix can be found by using the formula for a vertical parabola: \[y = k - p,\]where \(k = -1\) and \(p = 2\).
  • Thus, the directrix is \(y = -1 - 2 = -3\).
The directrix is always located opposite the focus relative to the vertex, forming one part of the key geometry that defines a parabola. It contributes to the unique property of parabolas: the distance from any point on the parabola to the focus is equal to the perpendicular distance from the point to the directrix, a property that is fundamental to understanding the parabola's geometry and behavior.

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

Spiral Path of a Dog \(\mathrm{A}\) dog is tied to a circular tree trunk of radius 1 \(\mathrm{ft}\) by a long leash. He has managed to wrap the entire leash around the tree while playing in the yard, and finds himself at the point \((1,0)\) in the figure. Seeing a squirrel, he runs around the tree counterclockwise, keeping the leash taut while chasing the intruder. (a) Show that parametric equations for the dog's path (called an involute of a circle) are \(x=\cos \theta+\theta \sin \theta \quad y=\sin \theta-\theta \cos \theta\) \([\text {Hint: Note that the leash is always tangent to the tree, }\) \(\text { so } O T \text { is perpendicular to } T D .]\) (b) Graph the path of the dog for \(0 \leq \theta \leq 4 \pi\)

\begin{array}{l}{1-22 \text { a pair of parametric equations is given. }} \\\ {\text { (a) Sketch the curve represented by the parametric equations. }} \\\ {\text { (b) Find a rectangular-coordinate equation for the curve by }} \\\ {\text { eliminating the parameter. }}\end{array} $$ x=\cos ^{3} t, \quad y=\sin ^{3} t, \quad 0 \leq t \leq 2 \pi $$

Let \(Z, Z^{\prime},\) and \(R\) be the matrices $$Z=\left[\begin{array}{l}{x} \\ {y}\end{array}\right] \quad Z^{\prime}=\left[\begin{array}{l}{X} \\ {Y}\end{array}\right]$$ $$R=\left[\begin{array}{cc}{\cos \phi} & {-\sin \phi} \\ {\sin \phi} & {\cos \phi}\end{array}\right]$$ Show that the Rotation of Axes Formulas can be written as $$Z=R Z^{\prime} \quad \text { and } \quad Z^{\prime}=R^{-1} Z$$

Lunar Orbit For an object in an elliptical orbit around the moon, the points in the orbit that are closest to and farthest from the center of the moon are called perilune and apolune, respectively. These are the vertices of the orbit. The center of the moon is at one focus of the orbit. The Apollo 11 spacecraft was placed in a lunar orbit with perilune at 68 \(\mathrm{mi}\) and apolune at 195 \(\mathrm{mi}\) above the surface of the moon. Assuming the moon is a sphere of radius 1075 \(\mathrm{mi}\) , find an equation for the orbit of Apollo \(11 .\) (Place the coordinate axes so that the origin is at the center of the orbit and the foci are located on the \(x\) -axis.)

Find parametric equations for the line with the given properties. Passing through \((12,7)\) and the origin
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

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.