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Chapter 10: Problem 7

Name the conic or limiting form represented by the given equation. Usually you will need to use the process of completing the square. $$ y^{2}-5 x-4 y-6=0 $$

Short Answer

Expert verified
The given equation represents a parabola.

Step by step solution

01

Rearrange the Equation

Start with the given equation \( y^2 - 5x - 4y - 6 = 0 \). Isolate terms involving \( y \) on one side: \( y^2 - 4y = 5x + 6 \).
02

Complete the Square for y

To complete the square for \( y \), take half the coefficient of \( y \) (which is -4), square it, and add it to both sides. Half of -4 is -2, and \((-2)^2 = 4\). Add 4 to both sides: \( (y^2 - 4y + 4) = 5x + 10 \).
03

Simplify the Equation

The left side becomes a perfect square: \((y - 2)^2 = 5x + 10 \).
04

Rearrange into Standard Form

Subtract 10 from both sides to get \( (y - 2)^2 = 5x + 10 - 10 \), which simplifies to \( (y - 2)^2 = 5x \).
05

Identify the Conic Section

The equation \((y - 2)^2 = 5x \) is in the form \( (y-k)^2 = 4p(x-h) \), which is the standard form of a parabola. Thus, the conic section is a parabola.

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Key Concepts

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

Completing the Square
Completing the square is a useful algebraic technique that transforms a quadratic equation into a form that is easier to work with.
For quadratic expressions resembling (\(ax^2 + bx + c = 0\)), completing the square involves creating a perfect square trinomial from the quadratic terms. This helps to easily solve or manipulate the equation.
  • Start by rearranging terms to isolate the quadratic and linear terms on one side of the equation.
  • Look at the coefficient of the linear term (the term with only a single variable). Take half of this coefficient, which provides the magic number needed for completing the square.
  • Square this magic number and add it to both sides of the equation. This creates a perfect square trinomial on one side, turning it into a binomial squared.
For example, in the expression \(y^2 - 4y\), half of \(-4\) is \(-2\), and \((-2)^2 = 4\). Thus, \((y^2 - 4y + 4)\) becomes \((y-2)^2\). This form makes it more straightforward to manipulate and solve equations.
Parabola
A parabola is one of the simplest yet fascinating figures in conic sections. It represents the path followed by a projectile under uniform gravitational force, among other real-world applications.
The parabola's shape is defined as the set of all points that are equidistant from a fixed point (called the "focus") and a line (called the "directrix").
  • Unlike circles, parabolas have an open curve structure.
  • They are typically represented algebraically by the vertex form \(y = a(x-h)^2 + k\) or \((y-k)^2 = 4p(x-h)\).
Parabolas can open upward, downward, left, or right, depending on their orientation in the equation. In our given exercise, the orientation equates to \((y-2)^2 = 5x\), representing a parabola that opens to the right.
Standard Form of a Parabola
The standard form of a parabola helps to quickly identify its essential features, such as the vertex, axis of symmetry, and the direction of opening. Generally represented as y = a(x-h)^2 + k for vertical parabolas and(x-h)^2 = 4p(y-k) for horizontal parabolas,
it simplifies the process of analyzing parabolic equations.
  • The vertex is the point \((h, k)\) where the parabola changes direction.
  • The parameter 4p describes the distance from the vertex to the focus as well as to the directrix, allowing us to determine how "wide" or "narrow" the parabola appears.
In our case, the equation \((y-2)^2 = 5x\) effectively transforms into \((y-k)^2 = 4p(x-h)\) with \(h = 0\), \(k = 2\), and \(p = 1.25\), confirming a standard form of a horizontal parabola.

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Most popular questions from this chapter

Find the area of the surface generated by revolving the curve \(x=t+\sqrt{7}, y=t^{2} / 2+\sqrt{7} t\), for \(-\sqrt{7} \leq t \leq \sqrt{7}\) about the \(y\)-axis.

Plot the points whose polar coordinates are \(\left(3, \frac{1}{3} \pi\right)\), \(\left(1, \frac{1}{2} \pi\right),\left(4, \frac{1}{3} \pi\right),(0, \pi),(1,4 \pi),\left(3, \frac{11}{7} \pi\right),\left(\frac{5}{3}, \frac{1}{2} \pi\right)\), and \((4,0)\).

Consider the ellipse \(x^{2} / a^{2}+y^{2} / b^{2}=1\). (a) Show that its perimeter is $$ P=4 a \int_{0}^{\pi / 2} \sqrt{1-e^{2} \cos ^{2} t} d t $$ where \(e\) is the eccentricity. C (b) The integral in part (a) is called an elliptic integral. It has been studied at great length, and it is known that the integrand does not have an elementary antiderivative, so we must turn to approximate methods to evaluate \(P\). Do so when \(a=1\) and \(e=\frac{1}{4}\) using the Parabolic Rule with \(n=4\). (Your answer should be near \(2 \pi\). Why?) (c) Repeat part (b) using \(n=20\).

In Problems 23-36, name the curve with the given polar equation. If it is a conic, give its eccentricity. Sketch the graph. \(r=\frac{6}{4-\cos \theta}\)

Show that \(\left(\sqrt{x^{2}-a^{2}}-x\right) \rightarrow 0\) as \(x \rightarrow \infty\).
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