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Consider the second-degree equation $$A x^{2}+C y^{2}+D x+E y+F=0$$ where \(A\) and \(C\) are not both \(0 .\) Show by completing the square: (a) If \(A C>0\), then the equation represents an ellipse, a circle, a point, or has no graph. (b) If \(A C<0\), then the equation represents a hyperbola or a pair of intersecting lines. (c) If \(A C=0\), then the equation represents a parabola, a pair of parallel lines, or has no graph.

Short Answer

Expert verified
AC > 0: Ellipse/circle/point or no graph; AC < 0: Hyperbola/intersecting lines; AC = 0: Parabola/parallel lines or no graph.

Step by step solution

01

Analyzing the equation

The given equation is a general second-degree equation of the form \(Ax^2 + Cy^2 + Dx + Ey + F = 0\). Our goal is to determine what type of conic section this equation represents, categorized by the conditions on \(A\), \(C\), and \(AC\).
02

Completing the square for x-term and y-term

First, focus on the \(x\)-terms of the equation. Rewrite \(Ax^2 + Dx\) as \(A(x^2 + \frac{D}{A}x)\). Completing the square results in \(A((x + \frac{D}{2A})^2 - \frac{D^2}{4A^2})\). Similarly, rewrite the \(y\)-terms \(Cy^2 + Ey\) as \(C(y^2 + \frac{E}{C}y)\), and completing the square results in \(C((y + \frac{E}{2C})^2 - \frac{E^2}{4C^2})\).
03

Simplifying constants

Adjust the equation to incorporate these completed squares. You'll have terms like \(Ax^2 + Dx\) and \(Cy^2 + Ey\) yielding to \(A(x + \frac{D}{2A})^2 - \frac{AD^2}{4A^2}\) and \(C(y + \frac{E}{2C})^2 - \frac{CE^2}{4C^2}\), respectively. Combining all, rewrite as: \[A(x + \frac{D}{2A})^2 + C(y + \frac{E}{2C})^2 = \frac{D^2}{4A} + \frac{E^2}{4C} - F\].
04

Case Analysis for Ellipse, Circle, Point, or No Graph

For \(AC > 0\), both \(A\) and \(C\) share the same sign, indicating the graph is an ellipse or a degenerate case: a circle when \(A = C\), a point if the equation simplifies to a single \(= 0\), or has no graph if inconsistently negative.
05

Case Analysis for Hyperbola or Pair of Intersecting Lines

For \(AC < 0\), \(A\) and \(C\) having different signs results in a hyperbola or potentially a pair of intersecting lines if the equation simplifies to a form with two linear equations.
06

Case Analysis for Parabola, Parallel Lines, or No Graph

When \(AC = 0\), one of the coefficients \(A\) or \(C\) is zero, resulting in a parabolic equation form. This could also represent parallel lines if the equation reduces to parallel line expressions, or it could have no graph if the equation resolves inconsistently.

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Key Concepts

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

Ellipse
An ellipse is one of the most common types of conic sections. When you encounter a second-degree equation like \(Ax^2 + Cy^2 + Dx + Ey + F = 0\) with the condition \(AC > 0\), you can determine the representation might be an ellipse. In this equation, both \(A\) and \(C\) share the same sign, indicating a similar scaling in both \(x\) and \(y\) directions.

The process of discovering this ellipse involves completing the square for both \(x\) and \(y\) terms. This method rewrites the equation so that it takes on a recognizable form. After completion, the equation can be observed as something akin to \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). Here, \((h,k)\) is the center of the ellipse, and \(a\) and \(b\) are the lengths of the semi-major and semi-minor axes, respectively.

Ellipse is shaped like an elongated circle and, contrary to a circle, isn’t confined to being perfectly round. If \(A = C\), the ellipse is, in fact, a circle. Degenerate forms of ellipses could also surface, presenting as either a single point or having no graph at all.
Hyperbola
In the case where the condition \(AC < 0\) holds for a second-degree equation, the equation is typically representing a hyperbola. Unlike the ellipse, the terms \(A\) and \(C\) have different signs, indicating opposite scaling along the axes.

Completing the square again transforms the equation into a familiar form, specifically \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) or \(-\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\). This demonstrates the classic hyperbola, where \((h,k)\) symbolizes the center, and the interaction depicts two distinct, opposing curves.

A hyperbola opens either horizontally or vertically, and appears to be mirror images across both axes. Additionally, when you have a degenerate form of this equation, it can simplify to a representation of a pair of intersecting lines, where the curves of the hyperbola appear to have met rather than plotted separately.
Parabola
Parabolas emerge in second-degree equations when \(AC = 0\). This condition signifies that either \(A\) or \(C\) is zero, indicating there’s no quadratic term in one of the variables, which gives the parabolic shape.

When you rewrite by completing the square, parabolas often take on a familiar form like \(y = ax^2 + bx + c\) or \(x = ay^2 + by + c\). This makes its graphical representation a single, continuous curve that can open upwards, downwards, left, or right.

In addition to representing a single parabola, the same condition \(AC = 0\) could yield a pair of parallel lines instead. This situation arises if, upon simplifying, the equation splits symmetrically. Another potential outcome is an equation with no graph, indicating it does not maintain a valid geometrical interpretation.

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Most popular questions from this chapter

Prove: The line tangent to the hyperbola $$ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $$ at the point \(\left(x_{0}, y_{0}\right)\) has the equation $$ \frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1 $$

(a) What is the area that is enclosed by one petal of the rose \(r=a \cos n \theta\) if \(n\) is an even integer? (b) What is the area that is enclosed by one petal of the rose \(r=a \cos n \theta\) if \(n\) is an odd integer? (c) Use a CAS to show that the total area enclosed by the rose \(r=a \cos n \theta\) is \(\pi a^{2} / 2\) if the number of petals is even. [Hint: See Exercise 78 of Section \(10.2 .]\) (d) Use a CAS to show that the total area enclosed by the rose \(r=a \cos n \theta\) is \(\pi a^{2} / 4\) if the number of petals is odd.

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Use the following values, where needed: radius of the Earth \(=4000 \mathrm{mi}=6440 \mathrm{~km}\) year (Earth year) \(=365\) days (Earth days) \(1 \mathrm{AU}=92.9 \times 10^{6} \mathrm{mi}=150 \times 10^{6} \mathrm{~km}\) The Hale-Bopp comet, discovered independently on July 23, 1995 by Alan Hale and Thomas Bopp, has an orbital eccentricity of \(e=0.9951\) and a period of 2380 years. (a) Find its semimajor axis in astronomical units (AU). (b) Find its perihelion and aphelion distances. (c) Choose a polar coordinate system with the center of the Sun at the pole, and find an equation for the Hale-Bopp orbit in that coordinate system. (d) Make a sketch of the Hale-Bopp orbit with reasonably accurate proportions.

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