Problem 3 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

Vertices are (0, 4) and (0, -4); foci are (0, 5) and (0, -5).

Step 1: Identify the Equation's Structure

The equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) represents a hyperbola that opens up and down. This is due to the \(y^2\) term being first and positive.

Step 2: Determine Standard Features from Equation

In the equation \(\frac{y^{2}}{4^{2}} - \frac{x^{2}}{3^{2}} = 1\), since it matches the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), it confirms the orientation of the hyperbola as vertical, where \(a = 4\) and \(b = 3\).

Step 3: Locate Vertices

For a vertical hyperbola, the vertices are located at \((0, \pm a)\). Thus, substituting \(a = 4\), the vertices are \((0, 4)\) and \((0, -4)\).

Step 4: Calculate Foci

The formula for foci of a hyperbola is \((0, \pm c)\), where \(c = \sqrt{a^2 + b^2}\). Calculate:\[c = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5\]Therefore, the foci are at \((0, 5)\) and \((0, -5)\).

Key Concepts

Vertices of a Hyperbola

Vertices are crucial points in understanding the geometry and location of a hyperbola on a graph. They represent points where the hyperbola approaches the axes most closely. For a vertical hyperbola, which is the one we are examining, the equation is given by:

• \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

In this form, the vertices will be found at \((0, \pm a)\). This is because the hyperbola opens up and down along the y-axis, due to the positive \(y^2 \) term. Substitute the specific value of \( a \):

• Here, \( a = 4 \), so the vertices are at \((0, 4)\) and \((0, -4)\).

This illustrates the maximum vertical stretch of the hyperbola from the center (0,0). It is important to remember that the distance between the vertices can indicate how "wide" the opening of the hyperbola is.

Foci of a Hyperbola

The foci of a hyperbola are two points located along its axis of symmetry and are used to define and describe the characteristics of the hyperbola. In a vertical hyperbola, they lie on the y-axis at the points \((0, \pm c)\). These points are vital because they help in describing how far the hyperbola's branches are "stretched" from the center.To find the foci, we use the formula:

• \(c = \sqrt{a^2 + b^2}\)

This formula stems from the equation’s inherent geometric properties, where \(a\) and \(b\) are constants that determine the scale of the hyperbola's axes:

• Using values from our equation, where \( a = 4 \) and \( b = 3 \), calculate \( c \): \( c = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \)
• This calculation gives us the foci at \((0, 5)\) and \((0, -5)\).

These foci are always located outside the vertices in a vertical hyperbola, which helps guide the shape of its curve.

Equations of Hyperbolas

Hyperbolas have distinct characteristics that differentiate them from other conic sections, such as ellipses and circles. The equation form gives insight into the orientation and geometry of the hyperbola. The type of hyperbola can be determined by the structure of its equation:

• A vertical hyperbola, like the one in this exercise, is represented by the equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
• Conversely, a horizontal hyperbola would be formed by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

The main elements \(a\) and \(b\) are indicators of the hyperbola's shape and orientation:

• \(a\) defines the distance from the center to each vertex along the axis that the hyperbola opens onto.
• \(b\) relates to the axis perpendicular to \(a\), impacting the width of each branch.

In our equation, knowing \( a = 4 \) and \( b = 3 \), the hyperbola opens vertically with a stretch of 4 units above and below the center, while maintaining symmetry about the y-axis. This symmetry influences the graph, accentuating its specific shape and orientation.