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Chapter 0: Problem 35

Plot the graph of the given equation. $$x^{2}-y^{2}=16$$

Short Answer

Expert verified
The graph is a hyperbola centered at (0,0) with vertices at (卤4,0) and asymptotes \(y = x\) and \(y = -x\).

Step by step solution

01

Identify the type of equation

The equation given is in the form of \(x^2 - y^2 = 16\). This is a standard form of a hyperbola equation centered at origin, specifically with axes aligned with the coordinate axes.
02

Determine the orientation

The negative sign in front of \(y^2\) indicates that the hyperbola opens along the x-axis. To confirm this, notice that the term with the positive coefficient (\(x^2\)) determines the orientation; here, it opens left and right.
03

Find the vertices

Rewrite the equation in standard form by dividing through by 16: \[\frac{x^2}{16} - \frac{y^2}{16} = 1\]. This is equivalent to \[\frac{x^2}{16} - \frac{y^2}{16} = 1\]. The vertices are at \((\pm 4, 0)\) because the square root of 16 is 4.
04

Determine the asymptotes

The equations of the asymptotes are found using the formula \(y = \pm \frac{b}{a}x\) where \(a^2 = 16\) and \(b^2 = 16\). Thus, \(b/a = 1\). Therefore, the asymptotes are \(y = x\) and \(y = -x\).
05

Plot the hyperbola

Draw the x and y axes. Locate the center at the origin \((0,0)\). Plot the vertices at \((4,0)\) and \((-4,0)\). Draw the asymptotes as diagonals \(y = x\) and \(y = -x\). Sketch the hyperbola opening to the left and right, approaching the asymptotes but never touching them.

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Key Concepts

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

Equation Orientation
The orientation of a hyperbola is dictated by the arrangement of its terms in the equation. A hyperbola typically follows the general forms of either
  • \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
  • \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\)
If the \(x^2\) term has a positive coefficient and comes first, as in this case where the equation is \(x^2 - y^2 = 16\), the hyperbola opens along the x-axis. This configuration indicates the hyperbola stretches horizontally; its branches face left and right.
However, if the \(y^2\) term were positive, the formula would suggest a hyperbola opening along the y-axis, appearing vertically on a graph. Understanding the orientation helps visualize how the hyperbola will appear once plotted.
Vertices of Hyperbola
Vertices are critical points on the hyperbola and determine its size and initial direction. To find the vertices of a hyperbola in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), you focus on the two focal points along the transverse axis.
For the equation \(\frac{x^2}{16} - \frac{y^2}{16} = 1\), the term under \(x^2\) indicates the squared length of the semi-major axis, \(a^2 = 16\). Therefore, \(a = 4\).
This reveals the vertices are at \((\pm 4, 0)\). They symbolize where each arm of the hyperbola is closest to the origin, defining the points through which the curve transitions direction radically.
Asymptotes of Hyperbola
Asymptotes in hyperbolas are imaginary lines that the curve approaches but never quite touches. These lines provide a framework shaping the hyperbola's direction and openness.
The general approach to determining asymptotes involves the equation \(y = \pm \frac{b}{a}x\). For the hyperbola \(\frac{x^2}{16} - \frac{y^2}{16} = 1\), both \(a^2\) and \(b^2\) are 16, thus \(a = b = 4\).
Plugging the values in, \(\frac{b}{a} = 1\).
This results in asymptotic equations of \(y = x\) and \(y = -x\). The hyperbola will contour towards these diagonal lines, forming an inward dash where its branches expand away from the origin.

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Most popular questions from this chapter

Explain the mistake that is made. Solve the equation \(\sqrt{3 t+1}=-4\) Solution: \(3 t+1=16\) $$ \begin{aligned} 3 t &=15 \\ t &=5 \end{aligned} $$ This is incorrect. What mistake was made?

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