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Chapter 4: Problem 50

Identify and sketch the graph of the conic section. $$ y^{2}-6 y-4 x+21=0 $$

Short Answer

Expert verified
The conic is a parabola that opens to the right with the vertex at (0,3), focus at (1,3), and directrix at x=-1.

Step by step solution

01

Convert to Standard Form

Rearrange the equation \(y^{2}-6 y-4 x+21=0\) by grouping the y-terms together and moving the x-term to the other side. Then, complete the square. The result is \((y-3)^2 - 4x = 0\), or in standard form: \((y-3)^2 = 4x\).
02

Identify the Conic Section

In the standard form \((y-k)^2 = 4p(x-h)\), this is an equation of a parabola that opens right if p is positive. Here, k=3 and p=4, which suggest that the parabola is centered at (0,3) and opens to the right.
03

Plot the Conic Section

Plot the vertex of the parabola at (0,3). Since the parabola opens to the right, and 4p=4, p=1, so the focus is 1 unit to the right of the vertex, at (1,3). The directrix is a vertical line 1 unit to the left of the vertex, at x=-1.

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Key Concepts

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

Parabola Standard Form
The standard form of a parabolic equation plays a crucial role in identifying and graphing parabolas. It's a streamlined equation that makes it easy to pinpoint the parabola's key features: its vertex, axis of symmetry, direction of opening, and the distance between the vertex and focus (p).
A parabola鈥檚 standard form is either
For vertical parabolas: \[(y-k)^2 = 4p(x-h)\]
For horizontal parabolas: \[(x-h)^2 = 4p(y-k)\]
In these equations,
  • \((h, k)\) represents the vertex of the parabola.
  • \(p\) is the distance from the vertex to the focus and also from the vertex to the directrix.
  • If \(p\) is positive in a horizontal parabola, it opens to the right, while a negative \(p\) opens to the left. For vertical parabolas, a positive \(p\) indicates that the parabola opens upwards, and a negative \(p\) opens downwards.
Understanding the standard form will greatly simplify the process of sketching the graph of a parabola.
Completing the Square
Completing the square is a foundational technique in algebra, often employed for rewriting quadratic expressions to help solve quadratic equations and graph conic sections like parabolas.
To complete the square:
  • First, ensure that the quadratic term has a coefficient of 1. If it does not, factor out the leading coefficient.
  • Next, arrange the terms of the quadratic expression so that the constant term is on the opposite side of the equation.
  • Take half of the coefficient of the linear term, square it, and add it to both sides of the equation.
  • Factor the result on the side of the quadratic and linear terms to form a perfect square trinomial. Consequently, the equation assumes the form \[(a+b)^2\], which is simpler to analyze and solve.
In our case, \(y^2 - 6y\) is the expression we want to transform. Taking half of -6, we square it to get 9 and add it to both sides, resulting in \((y-3)^2\). This reveals the vertex of the parabola, making the equation much easier to understand and graph.
Conic Sections in Algebra
Conic sections are the curves obtained when a plane intersects a cone at different angles. In algebra, these shapes are represented by second-degree polynomial equations, and each conic section鈥攃ircles, ellipses, parabolas, and hyperbolas鈥攈as a distinct standard form that facilitates their study and graphing.
Identifying the type of conic section from its general equation involves analyzing the squared terms:
  • If only one variable is squared, you鈥檙e dealing with a parabola.
  • If both variables are squared with the same coefficients but opposite signs, it indicates a hyperbola.
  • If both variables are squared with the same coefficients and signs, you get a circle, and when the coefficients are different, it鈥檚 an ellipse.
Algebraic manipulation, such as completing the square, is often used to convert the general equation into one of these standard forms, allowing for easier analysis and graphing of the conic section.

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Most popular questions from this chapter

(a) verify that each solution satisfies the differential equation, (b) test the set of solutions for linear independence, and (c) if the set is linearly independent, then write the general solution of the differential equation. $$ y^{\prime \prime \prime}+4 y^{\prime}=0 \quad\\{1,2 \cos 2 x, 2+\cos 2 x\\} $$

Perform a rotation of axes to eliminate the xy-term, and sketch the graph of the conic. $$ 5 x^{2}-2 x y+5 y^{2}-24=0 $$

Proof Let \(\left\\{y_{1}, y_{2}\right\\}\) be a set of solutions of a second- order linear homogeneous differential equation. Prove that this set is linearly independent if and only if the Wronskian is not identically equal to zero.

Perform a rotation of axes to eliminate the xy-term, and sketch the graph of the conic. $$ x y-8 x-4 y=0 $$

Determine whether the nonhomogeneous system \(A x=b\) is consistent. If it is, write the solution in the form \(\mathbf{x}=\mathbf{x}_{p}+\mathbf{x}_{\mathbf{k}},\) where \(\mathbf{x}_{\mathbf{p}}\) is a particular solution of \(\mathbf{A} \mathbf{x}=\mathbf{b}\) and \(x_{k}\) is a solution of \(A x=0\) $$ \begin{array}{c} x+2 y-4 z=-1 \\ -3 x-6 y+12 z=3 \end{array} $$
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