Problem 41 - Calculus Early Transcendentals

Question

Sketch a graph of the following hyperbolas. Specify the coordinates of the vertices and foci, and find the equations of the asymptotes. Use a graphing utility to check your work. $$4 x^{2}-y^{2}=16$$

Step-by-Step Solution

Answer

Question: Sketch the hyperbola with the equation $$4x^2 - y^2 = 16$$, and find its vertices, foci, and asymptotes. Answer: The vertices of the hyperbola are at \((-2, 0)\) and \((2, 0)\), the foci are at \((-\sqrt{20}, 0)\) and \((\sqrt{20}, 0)\), and the asymptotes have the equations \(y = 2x\) and \(y = -2x\).

Step 1: Write the hyperbola in standard form

To get the standard formula, rewrite the equation $$4x^2 - y^2 = 16$$. We can divide both sides by 16 to get the equation in standard form: $$\frac{x^2}{4} - \frac{y^2}{16} = 1$$ Thus, the standard form of the given hyperbola is: $$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$ with \(a = 2\), \(b = 4\), \(h = 0\), and \(k = 0\).

Step 2: Find the vertices and foci

The hyperbola's vertices are located at \((\pm a, 0)\) from the center \((h, k)\). Since we have \(h = 0\), \(k = 0\), and \(a = 2\), the vertices are at \((\pm 2, 0)\) or \((-2, 0)\) and \((2, 0)\). The foci are located at \((\pm \sqrt{a^2 + b^2}, 0)\). With \(a = 2\), \(b = 4\), and the origin as the center, the foci are at: $$(\pm \sqrt{2^2 + 4^2}, 0) = (\pm \sqrt{20}, 0)$$

Step 3: Find the equations of the asymptotes

The asymptotes have the equations: $$y = \pm \frac{b}{a} (x - h) + k$$ Substituting \(a = 2\), \(b = 4\), \(h = 0\), and \(k = 0\), we get the equations of the asymptotes: $$y = \pm 2x$$

Step 4: Sketch the hyperbola and check with a graphing utility

Now, we can sketch the hyperbola. First, plot the center of the hyperbola (0,0), the vertices \((-2,0)\) and \((2,0)\), and draw the asymptotes with the equations \(y = 2x\) and \(y = -2x\). Then, sketch the hyperbola, making sure it approaches the asymptotes as it extends. Finally, use a graphing utility to check if the sketch is accurate. To summarize, we have: 1. Vertices: \((-2, 0)\) and \((2, 0)\) 2. Foci: \((-\sqrt{20}, 0)\) and \((\sqrt{20}, 0)\) 3. Asymptotes: \(y = 2x\) and \(y = -2x\)

Key Concepts

Standard Form of Hyperbola

Converting a hyperbola’s equation into its standard form is the first essential step in understanding its properties. The general standard form of a hyperbola is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]when it opens horizontally. In contrast, for a hyperbola that opens vertically, the equation would be \[ \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \].

It’s important to note the positioning of the terms: a hyperbola that opens with a horizontal asymptote will have the \((x-h)^2\) term first. In our specific exercise format \[ 4x^2 - y^2 = 16 \], dividing through by 16 gives us \[ \frac{x^2}{4} - \frac{y^2}{16} = 1 \]. This tells us several things:

• The center of the hyperbola is at \((h, k) = (0, 0)\)
• The value \(a^2 = 4\) hence \(a = 2\) tells us the distance from the center to each vertex.
• The value \(b^2 = 16\) hence \( b = 4\) is used in determining the distance of the foci and the asymptotes.

Vertices of Hyperbola

The vertices of a hyperbola are key points that show the hyperbola's widest point along its transverse axis. These points are crucial in sketching the curve accurately.

For a horizontally oriented hyperbola (like in this exercise), the vertices are found using the formula \((h \pm a, k)\). From our standard form \(\frac{x^2}{4} - \frac{y^2}{16} = 1\):

• \( h = 0\), \( k = 0\) - so the center is at the origin.
• \( a = 2 \) - which means vertices are \((-2, 0) \) and \((2, 0)\).

These coordinates indicate where the hyperbola begins to curve back inwards and are vital in plotting. The vertices not only define the shape but also help in determining the direction the hyperbola opens, which in this case is horizontally.

Foci of Hyperbola

The foci of a hyperbola are another critical set of points that lie on the transverse axis, further out than the vertices. They play a fundamental role in defining the hyperbola's shape. To find the foci, we use the relation \(\sqrt{a^2+b^2}\). For our hyperbola:

• \(a^2 = 4\)
• \(b^2 = 16\)

So, the distance to each focus from the center is \(\sqrt{4 + 16} = \sqrt{20}\). This gives us focus positions at \((\pm \sqrt{20}, 0)\). The foci reflect how 'stretched' the hyperbola is and more technically, define the hyperbola by the property that the difference in distances from any point on the hyperbola to the foci is constant. Thus, no sketch is complete without marking the foci.

Asymptotes of Hyperbola

Asymptotes are the guidelines that the hyperbola approaches but never actually reaches as it stretches infinitely. For our horizontal hyperbola, the asymptotes are the diagonals of a rectangle centered on \((h, k)\) with dimensions \(2a\) by \(2b\).

To express them mathematically, use the formula: \(y = \pm \frac{b}{a} (x - h) + k\) with \(a = 2\), \(b = 4\), \(h = 0\), and \(k = 0\). Substituting these values, we get:

• \( y = \pm 2x \)

These linear equations guide us in sketching the hyperbola correctly. The lines never touch the curve but show its directional approach as it stretches horizontally and vertically. They are important as a visual tool when sketching by hand and help affirm the symmetry and bounds of the hyperbola.