Problem 10 - Precalculus: Functions and Graphs

Question

Find the vertices, the foci, and the equations of the asymptotes of the hyperbola. Sketch its graph, showing the asymptotes and the foci. $$y^{2}-16 x^{2}=1$$

Step-by-Step Solution

Answer

Vertices: (0, 1) and (0, -1); Foci: (0, ±√17/4); Asymptotes: y = ±4x.

Step 1: Standard Form of Hyperbola

To find the vertices, foci, and asymptotes of a hyperbola, we first put the equation in its standard form. The given equation is \( y^2 - 16x^2 = 1 \), which is already in the standard form \( \frac{y^2}{1} - \frac{x^2}{\left(\frac{1}{4}\right)} = 1 \) for a vertically oriented hyperbola centered at the origin \((0,0)\). Here, \( a^2 = 1 \) so \( a = 1 \), and \( b^2 = \frac{1}{16} \) so \( b = \frac{1}{4} \).

Step 2: Finding the Vertices

The vertices of a hyperbola with a vertical transverse axis are located at \((0, \pm a)\) where \( a = 1 \). Thus, the vertices are at \( (0, 1) \) and \( (0, -1) \).

Step 3: Finding the Foci

The foci of a hyperbola are calculated using the formula \( c = \sqrt{a^2 + b^2} \). Plugging in the values, we get \( c = \sqrt{1 + \left(\frac{1}{16}\right)} = \sqrt{\frac{17}{16}} = \frac{\sqrt{17}}{4} \). Therefore, the foci are located at \( (0, \pm \frac{\sqrt{17}}{4}) \).

Step 4: Equations of the Asymptotes

For a hyperbola with a vertical transverse axis, the equations of the asymptotes are given by \( y = \pm \frac{a}{b} x \). With \( a = 1 \) and \( b = \frac{1}{4} \), the equations become \( y = \pm 4x \).

Step 5: Sketching the Graph

To sketch the graph, plot the vertices at (0,1) and (0,-1), and the foci at \((0, \pm \frac{\sqrt{17}}{4})\). Draw the asymptotes, which are the lines y = 4x and y = -4x. Sketch the hyperbola branches opening upwards and downwards along the y-axis and approaching these asymptotes.

Key Concepts

Vertices of Hyperbola

To find the vertices of the hyperbola, it's important to understand that these are the points where the hyperbola intersects its principal axis. For the equation \( y^2 - 16x^2 = 1 \), the hyperbola is centered at the origin, and it's oriented vertically because the \( y^2 \) term is positive. This means the transverse axis, or the main axis running through the center along which the hyperbola opens, is vertical.

• Since the standard form for a vertical hyperbola is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the vertices are located at \( (0, \pm a) \).
• For this equation, \( a^2 = 1 \), hence \( a = 1 \).
• The vertices are then found at \( (0, 1) \) and \( (0, -1) \).

These are the highest points along the y-axis where the hyperbola stretches before moving out towards infinity.

Foci of Hyperbola

The foci of a hyperbola are crucial as they determine the shape and dimensions of the hyperbola's curves. They lie along the transverse axis, further out than the vertices.

• For the hyperbola \( y^2 - 16x^2 = 1 \), calculating the foci involves the term \( c = \sqrt{a^2 + b^2} \).
• With \( a^2 = 1 \) and \( b^2 = \frac{1}{16} \), it follows that \( c = \sqrt{1 + \frac{1}{16}} = \frac{\sqrt{17}}{4} \).
• This gives us the foci at \( (0, \pm \frac{\sqrt{17}}{4}) \).

The foci are points towards which the hyperbola curves converge. These are outside and further from the center than the vertices, lying along the y-axis.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never actually touches. These guide the hyperbola's shape and provide a visual limit to its extensions. For a vertical hyperbola like \( y^2 - 16x^2 = 1 \), the asymptotes provide a framework.

• The general form for determining the equations of asymptotes for a hyperbola is \( y = \pm \frac{a}{b} x \).
• Since \( a = 1 \) and \( b = \frac{1}{4} \), the slopes of the asymptotes are \( \pm 4 \).
• Thus, the equations are \( y = 4x \) and \( y = -4x \).

These lines are critical to drawing the typically mirrored "bow" shapes that the branches of the hyperbola form.

Equation of Hyperbola

The equation of the hyperbola provides all the necessary pieces to understand its shape, orientation, and dimensions. A hyperbola can have either a vertical or horizontal orientation, based on which variable is positive in the formula.

• The given equation is \( y^2 - 16x^2 = 1 \).
• This corresponds to the standard form of a vertical hyperbola because the \( y^2 \) term is greater, indicating the transverse axis is along the y-axis.
• This shows us the center is at the origin \((0,0)\) with \( a = 1 \) and \( b = \frac{1}{4} \) derived from their respective squares \((a^2 = 1, b^2 = \frac{1}{16})\).

This form is critical for identifying other elements such as vertices, foci, and asymptotes.

Graphing Hyperbola

Graphing a hyperbola might seem complex, but breaking it into steps can simplify the process. With the equation \( y^2 - 16x^2 = 1 \), the focus is on displaying its main components accurately in a sketch.

• First, plot the vertices \((0, 1)\) and \((0, -1)\) on the graph.
• Next, draw the foci at \((0, \pm \frac{\sqrt{17}}{4})\) to show the focal points of curvature.
• The asymptotes, \( y = 4x \) and \( y = -4x \), should be lightly penciled in to provide a guide for the curve's approach.
• Finally, sketch the hyperbola's branches. Since it's vertical, the arms will open along the y-axis, swooping towards but never touching the asymptotes.

This comprehensive approach ensures a clear visual representation, connecting all calculated elements of the hyperbola effectively.