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Chapter 11: Problem 2

The graph of the equation \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) with \(a>0, b>0\) is a hyperbola with _______________ (horizontal/vertical) transverse axis, vertices (___, ___) and (___, ___) and foci \((\pm c, 0),\) where \(c=\) _______________ . So the graph of \(\frac{x^{2}}{4^{2}}-\frac{y^{2}}{3^{2}}=1\) is a hyperbola with vertices (___, ___) and (___, ___) and foci (___, ___) and (___, ___).

Short Answer

Expert verified
It's a horizontal hyperbola with vertices (4, 0), (-4, 0), and foci (5, 0), (-5, 0).

Step by step solution

01

Identify the type of hyperbola

The equation \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \) represents a hyperbola with a horizontal transverse axis because the term with \( x^2 \) is positive.
02

Determine the vertices

The vertices of the hyperbola are located at \((\pm a, 0)\). Given \( a^2 = 4^2 \), we find that \( a = 4 \). Thus, the vertices are \( (4, 0) \) and \( (-4, 0) \).
03

Calculate the foci using c

The distance to the foci \( c \) is calculated using the formula \( c^2 = a^2 + b^2 \). Here, \( a^2 = 16 \) and \( b^2 = 9 \), so \( c^2 = 16 + 9 = 25 \). Thus, \( c = \sqrt{25} = 5 \).
04

Determine the coordinates of the foci

With \( c = 5 \) and a horizontal transverse axis, the foci are at \((\pm c, 0)\), or \((5, 0)\) and \((-5, 0)\).

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Key Concepts

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

Transverse Axis
When we talk about a hyperbola, the transverse axis is a crucial concept. It refers to the axis around which the hyperbola is centered. Specifically, it's the line that passes through the center of the hyperbola and its two vertices.
In the equation \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \), the transverse axis is horizontal. This is determined by the positive \( x^2 \) term. Whenever \( x^2 \) appears before \( y^2 \), and with a positive coefficient, the hyperbola stretches out horizontally.
Understanding the transverse axis helps us immediately identify the orientation of the hyperbola on a graph:
  • For \( x^2 \) positive first, the transverse axis is horizontal.
  • If \( y^2 \) were positive first, the transverse axis would be vertical.
Vertices and Foci
Vertices and foci are key features that help define the shape and position of a hyperbola. The vertices are the points located at the endpoints of the hyperbola's transverse axis.
In our equation, the vertices are found using \((\pm a, 0)\). Since \( a^2 = 4^2 \), \( a \) equals 4, giving vertices at \((4, 0)\) and \((-4, 0)\). Simplifying the idea:
  • The vertices are always symmetrically located along the transverse axis at \( \pm a \).
For the foci, they lie further out than the vertices along the transverse axis. Calculate them using the formula \( c^2 = a^2 + b^2 \). Here, \( c^2 = 16 + 9 = 25 \), so \( c = 5 \). Therefore, the foci are \((5, 0)\) and \((-5, 0)\).
  • The foci help define how "stretched" the hyperbola is.
  • They always lie outside the vertices.
Mathematical Equations
A hyperbola is defined by a specific type of mathematical equation. In its standard form, it is represented as \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \). This equation tells us that we have a hyperbola with a horizontal transverse axis.
Let's break down this equation:
  • The \( \frac{x^{2}}{a^{2}} \) term signifies the horizontal stretch. It means our hyperbola extends further left and right.
  • The \( \frac{y^{2}}{b^{2}} \) part implies a vertical constraint. It minimizes the upward and downward spread compared to the horizontal one.
The presence of "1" on the right side ensures that the graph is a hyperbola, not another conic section.
To find the graph's key points, we often use these equations:
  • Vertices: \((\pm a, 0)\)
  • Foci: Use \( c^2 = a^2 + b^2 \) to calculate and find \((\pm c, 0)\)
Understanding these equations is fundamental to graphing and analyzing any hyperbola in mathematics.

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Most popular questions from this chapter

An equation of a parabola is given. (a) Find the focus, directrix, and focal diameter of the parabola. (b) Sketch a graph of the parabola and its directrix. $$y=-\frac{1}{8} x^{2}$$

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