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Chapter 5: Problem 43

Identify the graph of the given equation. $$3 x^{2}=y^{2}-1$$

Short Answer

Expert verified
The graph is a vertical hyperbola.

Step by step solution

01

Rearrange the Equation

The given equation is \(3x^2 = y^2 - 1\). To analyze this better, rearrange it to the standard form of a conic section. Add 1 to both sides to get \(3x^2 + 1 = y^2\). Now it becomes \(y^2 = 3x^2 + 1\).
02

Identify the Type of Conic Section

The standard form of a hyperbola equation is \(y^2/a^2 - x^2/b^2 = 1\) or \(x^2/a^2 - y^2/b^2 = 1\). By comparing \(y^2 = 3x^2 + 1\) with these standard forms, we can rewrite it as \(\frac{y^2}{1} - \frac{3x^2}{1} = 1\), which fits the form of \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).
03

Determine Characteristics of the Hyperbola

From \(\frac{y^2}{1} - \frac{3x^2}{1} = 1\), we see that this is a vertical hyperbola since the \(y^2\) term is positive and comes first. The center is at the origin \((0, 0)\), the transverse axis is along the y-axis, and the coordinates of the vertices depend on the specific values of \(a\) and \(b\), which we can compute if needed.

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Key Concepts

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

Conic Sections
Conic sections are curves obtained by intersecting a cone with a plane. These include ellipses, parabolas, circles, and hyperbolas. Their properties can be derived from their algebraic equations.
The different shapes depend on how the plane cuts through the cone:
  • Ellipse: When the plane cuts through the cone at an angle, but not parallel to the base, an ellipse is formed.
  • Circle: A special case of an ellipse, it occurs when the plane is perpendicular to the cone's axis.
  • Parabola: Formed when the plane is parallel to the side of the cone.
  • Hyperbola: Occurs when the plane cuts through both halves of the cone, generating two distinct curves.
Each conic section has a standard form equation. Recognizing these forms helps in identifying the graph type from its equation.
Equation Identification
Equation identification involves comparing a given equation to the standard forms of conic sections to determine its type. For hyperbolas, there are two forms:
One for horizontal hyperbolas: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), and another for vertical hyperbolas: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\).

When we rewrite an equation into one of these forms, the arrangement of terms reveals the conic section.
  • If the \(x^2\) term is positive and comes first, the hyperbola is horizontal.
  • If the \(y^2\) term is positive and comes first, the hyperbola is vertical.
The coefficients provide further information, such as the distances to the vertices and foci, which are essential for analyzing conic sections.
Vertical Hyperbola
A vertical hyperbola is a type of conic section where the transverse axis is vertical, aligning with the y-axis. Its equation in standard form is:
\(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Here, "a" and "b" are parameters that determine the hyperbola's size and shape.
For this equation:
  • The center of the hyperbola is at the origin, \((0,0)\).
  • The vertices are located at \((0,\pm a)\), indicating the extent of the hyperbola along the y-axis.
  • Asymptotes, crucial for sketching the hyperbola, are lines that guide its opening. For a vertical hyperbola, these are \(y = \pm\frac{a}{b}x\).
Recognizing a vertical hyperbola involves checking the order of \(y^2\) and \(x^2\) in the equation, ensuring \(y^2\) comes first to establish its orientation.

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Most popular questions from this chapter

Sometimes the graph of a quadratic equation is a straight line, a pair of straight lines, or a single point. We refer to such a graph as a degenerate conic. It is also possible that the equation is not satisfied for any values of the variables, in which case there is no graph at all and we refer to the conic as an imaginary conic. Identify the conic with the given equation as either degenerate or imaginary and, where possible, sketch the graph. $$x^{2}+2 x y+y^{2}=0$$

Evaluate the quadratic form \(f(\mathbf{x})=\mathbf{x}^{T} A \mathbf{x}\) for the given A and x. $$A=\left[\begin{array}{rrr} 1 & 0 & -3 \\ 0 & 2 & 1 \\ -3 & 1 & 3 \end{array}\right], \mathbf{x}=\left[\begin{array}{r} 2 \\ -1 \\ 1 \end{array}\right]$$

Find the orthogonal decomposition of v with respect to \(W\). $$\mathbf{v}=\left[\begin{array}{r} 2 \\ -2 \end{array}\right], W=\operatorname{span}\left(\left[\begin{array}{l} 1 \\ 3 \end{array}\right]\right)$$

Identify the conic with the given equation and give its equation in standard form. $$2 x y+2 \sqrt{2} x-1=0$$

Classify each of the quadratic forms as positive definite, positive semidefinite, negative definite, negative semidefinite, or indefinite. $$-x^{2}-y^{2}-z^{2}-2 x y-2 x z-2 y z$$
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