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Chapter 11: Problem 23

Graph the equations. $$4 x y+3 y^{2}+4 x+6 y=1$$

Short Answer

Expert verified
Express and plot using technology for clarity; analyze points and use potential solving transforms.

Step by step solution

01

Identify the Equation Type

The given equation is $4xy + 3y^2 + 4x + 6y = 1$. This is a quadratic equation in two variables, which suggests it is a conic section. However, its specific shape isn't obvious from the equation as written.
02

Manipulate Equation into Standard Form

To better understand the shape, complete the square if necessary. However, in this instance, solving for $y$ directly might provide better intuition or graphing technology might assist. Observing the mixed term $4xy$, it indicates a rotated conic. Completing the square in such situations without rotation can be less straightforward. Instead, solve for $y$ to explore easily plottable points.
03

Solve for Intersections with Axes

Set \(x = 0\) in the equation to find where it intersects the y-axis:\[3y^2 + 6y = 1 \Rightarrow 3y^2 + 6y - 1 = 0\]This is a quadratic equation in \(y\), apply the quadratic formula to find y-intercepts. Similarly, set \(y = 0\) to find x-intercepts:\[4x + 4x = 1 \Rightarrow 8x = 1 \Rightarrow x = \frac{1}{8}\]
04

Verify Intersection Points

For the y-intercepts, calculate using the quadratic formula:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\text{ where }a=3, b=6, c=-1\]Substitute to find:\[y = \frac{-6 \pm \sqrt{36 + 12}}{6} = \frac{-6 \pm \sqrt{48}}{6} = \frac{-6 \pm 4\sqrt{3}}{6}\]Thus, providing two y-intercepts if real roots exist.
05

Plotting Points and Exploring Nature

Consider the transformed forms or direct plotting using software to observe an approximate sketch of the curve. The method suggests points computed, and the curve might be plotted as such transformed functions indicating rotation or mixed terms potentially parabolic in rotated form.For axes intersections, plot those specific points: \((0, \frac{-6 + 4\sqrt{3}}{6}), (0, \frac{-6 - 4\sqrt{3}}{6})\) and \(\left(\frac{1}{8}, 0\right)\). Plotting with software may yield a clearer depiction.
06

Graph Visualization and Additional Points

Using graphical software or additional algebraic manipulation incorporating parameters would give a better visualization. The mixed term could be reduced or expressed in rotated coordinates for clarity, especially for students encountering such transformations in extended algebra or pre-calculus. Combining transformed equations or using graphing techniques, visualize a curve fitting parabolic lines with the computed intersection points.

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Key Concepts

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

Quadratic Equations
Quadratic equations in two variables, such as the given equation \( 4xy + 3y^2 + 4x + 6y = 1 \), define curves known as conic sections. These equations can form parabolas, ellipses, hyperbolas, or even degenerate forms. Their general form is expressed as \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \), where the presence of the \( Bxy \) term suggests a rotated conic. This adds complexity because the axes of the conic are not aligned with the coordinate axes. Recognizing a quadratic equation helps determine the graphing method and identify potential transformations needed to visualize the conic correctly.
Completing the Square
Completing the square is a method used to convert a quadratic expression into a perfect square trinomial, which makes it easier to analyze or graph. However, in cases like our equation \( 4xy + 3y^2 + 4x + 6y = 1 \), the presence of the \( 4xy \) term complicates the process. When facing such rotated conics, completing the square might involve a change of variables or rotation of axes, which requires advanced algebra. In simpler problems, completing the square helps to rewrite the equation in a standard conic form, making interpretation and graphing easier. It breaks down complex expressions into parts that reveal the nature and dimensions of the conic section.
Intersections with Axes
Finding where the graph intersects the axes provides crucial anchor points for sketching conic sections. Intersections occur when either \( x = 0 \) or \( y = 0 \). For the given equation, setting \( x = 0 \) leads to the quadratic \( 3y^2 + 6y - 1 = 0 \), solved using the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Plugging the terms, we find potential y-intercepts. Similarly, by setting \( y = 0 \), we solve for \( x \, = \, \frac{1}{8} \), identifying the x-intercept. These intersections provide a framework for the curve's placement and guide further examination with graphing tools or detailed calculations.
Rotated Conics
A rotated conic occurs when the axes of symmetry of a conic section are not parallel to the coordinate axes, often due to a non-zero \( Bxy \) term in the equation. These conics are transformed by rotating the axes, requiring knowledge of trigonometric transformations. The rotation aligns the conic with the new axes, often simplifying the equation form. For comprehensive graphing, computers or software can rotate the graph, providing a clearer picture of the conic's true nature. Understanding rotated conics gives insight into more complex algebraic transformations and enhances the ability to recognize these forms in broader mathematical contexts. This helps to visualize and comprehend how rotation affects the position and orientation of conic sections.

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Most popular questions from this chapter

Graph the equations. $$(x+y)^{2}+4 \sqrt{2}(x-y)=0$$

A normal to an ellipse is a line drawn perpendicular to the tangent at the point of tangency. Show that the equation of the normal to the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1\) at the point \(\left(x_{1}, y_{1}\right)\) can be written $$a^{2} y_{1} x-b^{2} x_{1} y=\left(a^{2}-b^{2}\right) x_{1} y_{1}$$

Find the equation of the parabola satisfying the given conditions. In each case, assume that the vertex is at the origin. The parabola is symmetric about the \(x\) -axis, the \(x\) -coordinate of the focus is negative, and the length of the focal chord perpendicular to the \(x\) -axis is 9

In designing an arch, architects and engineers sometimes use a parabolic arch rather than a semicircular arch. (One reason for this is that, in general, the parabolic arch can support more weight at the top than can the semicircular arch.) In the following figure, the blue arch is a semicircle of radius \(1,\) centered at the origin. The red arch is a portion of a parabola. As is indicated in the figure, the two arches have the same base and the same height. Assume that the unit of distance for each axis is the meter. GRAPH CANT COPY (a) Find the equation of the parabola in the figure. (b) Using calculus, it can be shown that the area under this parabolic arch is \(\frac{4}{3} \mathrm{m}^{2} .\) Assuming this fact, show that the area beneath the parabolic arch is approximately \(85 \%\) of the area beneath the semicircular arch. (c) Using calculus, it can be shown that the length of this parabolic arch is \(\sqrt{5}+\frac{1}{2} \ln (2+\sqrt{5})\) meters. Assuming this fact, show that the length of the parabolic arch is approximately \(94 \%\) of the length of the semicircular arch.

Let \(\overline{P Q}\) be a focal chord of the parabola \(y^{2}=4 p x,\) and let \(M\) be the midpoint of \(\overline{P Q} .\) A perpendicular is drawn from \(M\) to the \(x\) -axis, meeting the \(x\) -axis at \(S .\) Also from \(M,\) a line segment is drawn that is perpendicular to \(\overline{P Q}\) and that meets the \(x\) -axis at \(T\). Show that the length of \(\overline{S T}\) is onehalf the focal width of the parabola.
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