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Chapter 10: Problem 46

Prove or find counterexamples to the following statements about the graph of \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) a. If \(A C>0\) , the graph is an ellipse. b. If \(A C>0\) , the graph is a hyperbola. c. If \(A C<0,\) the graph is a hyperbola.

Short Answer

Expert verified
a) True; b) False; c) True.

Step by step solution

01

Understanding the equation of the conic

The general second-degree equation \(A x^{2}+B x y+C y^{2}+D x+E y+F=0\) can represent an ellipse, hyperbola, parabola, or degenerate conic depending on the values of \(A, B,\) and \(C\). The type of conic depends on the discriminant \(\Delta = B^2 - 4AC\).
02

Discriminant condition for an ellipse

For the equation to represent an ellipse, the condition is \(B^2 - 4AC < 0\). Additionally, \(A\) and \(C\) need to have the same sign for the graph to truly be an ellipse; thus, \(AC > 0\). This means statement (a) is generally true.
03

Discriminant condition for a hyperbola

For the graph to be a hyperbola, the condition \(B^2 - 4AC > 0\) must hold. If \(AC > 0\), it is not possible for \(B^2 - 4AC\) to be positive since \(B^2 < 4AC\) in such cases; therefore, statement (b) cannot be true as it violates this requisite.
04

Condition for a hyperbola when AC < 0

When \(AC < 0\), it directly affects the discriminant \(B^2 - 4AC\) by making it positive, as \(B^2\) will be inherently greater than \(4AC\). Hence, when \(AC < 0\), \(B^2 - 4AC > 0\) making statement (c) true.

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Key Concepts

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

Ellipses
Ellipses are a type of conic section that appears as an oval shape. In mathematics, an ellipse can be described by a specific form of the second-degree equation: \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). For a graph to be classified as an ellipse, two important conditions must be met:

  • The discriminant \(B^2 - 4AC\) should be less than zero.
  • The coefficients \(A\) and \(C\) should have the same sign, which means \(AC > 0\).
These criteria ensure that the conic section opens as an ellipse. Ellipses have unique characteristics:
  • They are closed curves.
  • They have two foci, and the sum of the distances from any point on the ellipse to the foci is constant.
Understanding these properties helps us recognize why, when \(AC > 0\), the graph is an ellipse.
Hyperbolas
Hyperbolas are another type of conic section, creating two separate curves that open outwards. These curves can be defined by the same second-degree equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\), but with different conditions than those for ellipses. To classify a graph as a hyperbola, the following should be true:

  • The discriminant \(B^2 - 4AC\) must be greater than zero.
  • The equation often requires \(AC < 0\), meaning \(A\) and \(C\) have opposite signs.
These conditions cause the curves to open in opposite directions, forming the shape of a hyperbola. Hyperbolas display certain features:
  • They consist of two parts called branches.
  • Each branch gets infinitely close to a pair of intersecting lines known as asymptotes but never actually meets them.
With \(AC < 0\), it ensures the discriminant's positivity, further explaining why if \(AC < 0\), the graph represents a hyperbola.
Discriminant Condition
The discriminant condition plays a crucial role in determining the type of conic section represented by the equation \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\). This critical factor is known as the discriminant, represented by the formula \(B^2 - 4AC\).

  • If \(B^2 - 4AC < 0\), the graph is an ellipse.
  • If \(B^2 - 4AC > 0\), the graph is a hyperbola.
  • If \(B^2 - 4AC = 0\), the graph is a parabola or sometimes a degenerate conic.
The discriminant condition is instrumental for classifying any second-degree equation into its correct conic section category. This is because it takes into account the interplay between the coefficients \(A\), \(B\), and \(C\), governing the overall shape of the graph.
Second-Degree Equation
A second-degree equation in two variables is a powerful expression that can represent various conic sections. The general form \(Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0\) encompasses not just ellipses and hyperbolas, but also parabolas and potentially degenerate cases.

To understand these possibilities:
  • An ellipse forms when the discriminant \(B^2 - 4AC < 0\) and \(AC > 0\).
  • A hyperbola results from \(B^2 - 4AC > 0\) and \(AC < 0\).
  • A parabola occurs when \(B^2 - 4AC = 0\).
This equation type is pivotal because it defines every major shape in conic sections. Each shape 鈥� ellipse, hyperbola, or parabola 鈥� presents distinct geometric properties and graph characteristics. Recognizing the influence of the second-degree equation's coefficients on these forms unlocks a deeper understanding of geometry and algebra.

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Most popular questions from this chapter

In Exercises \(65-68\) , find a polar equation for the given curve. In each case, sketch a typical curve. $$ y^{2}=4 a x+4 a^{2} $$

Exercises \(29-36\) give the eccentricities of conic sections with one focus at the origin, along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$ e=1 / 5, \quad y=-10 $$

Exercises \(45-48\) give equations for parabolas and tell how many units up or down and to the right or left each parabola is to be shifted. Find an equation for the new parabola, and find the new vertex, focus, and directrix. $$ x^{2}=6 y, \quad \text { left } 3, \text { down } 2 $$

The reflective property of parabolas The figure here shows a typical point \(P\left(x_{0}, y_{0}\right)\) on the parabola \(y^{2}=4 p x .\) The line \(L\) is tangent to the parabola at \(P .\) The parabola's focus lies at \(F(p, 0) .\) The ray \(L^{\prime}\) extending from \(P\) to the right is parallel to the \(x\) -axis. We show that light from \(F\) to \(P\) will be reflected out along \(L^{\prime}\) by showing that \(\beta\) equals \(\alpha\) . Establish this equality by taking the following steps. a. Show that \(\tan \beta=2 p / y_{0}\) . b. Show that \(\tan \phi=y_{0} /\left(x_{0}-p\right)\) c. Use the identity $$\tan \alpha=\frac{\tan \phi-\tan \beta}{1+\tan \phi \tan \beta}$$ to show that \(\tan \alpha=2 p / y_{0}\) since \(\alpha\) and \(\beta\) are both acute, \(\tan \beta=\tan \alpha\) implies \(\beta=\alpha\)

Exercises \(29-36\) give the eccentricities of conic sections with one focus at the origin, along with the directrix corresponding to that focus. Find a polar equation for each conic section. $$ e=5, \quad y=-6 $$
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