Problem 2 - Precalculus Mathematics for Calculus

Question

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 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 (________,_______).

Step-by-Step Solution

Answer

Vertices: (4, 0) and (-4, 0); Foci: (5, 0) and (-5, 0).

Step 1: Identify the Hyperbola Structure

The equation \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) represents a hyperbola with a horizontal transverse axis. The vertices are located at \((\pm a, 0)\) and the foci at \((\pm c, 0)\), where \(c\) is calculated using \(c^{2} = a^{2} + b^{2}\).

Step 2: Substitute Given Values

Given \(\frac{x^{2}}{4^{2}}-\frac{y^{2}}{3^{2}}=1\), identify \(a=4\) and \(b=3\).

Step 3: Calculate the Vertices

The vertices will be at \((\pm a, 0)\), meaning they are at \((\pm 4, 0)\). Therefore, the vertices are \((4, 0)\) and \((-4, 0)\).

Step 4: Calculate the Foci

Using the formula \(c^{2} = a^{2} + b^{2}\), substitute the values to get \(c^{2} = 4^{2} + 3^{2} = 16 + 9 = 25\). Thus, \(c = \sqrt{25} = 5\). The foci are located at \((\pm c, 0)\), which means \((5, 0)\) and \((-5, 0)\).

Key Concepts

Vertices of Hyperbola

Understanding vertices is crucial when studying hyperbolas. The vertices of a hyperbola are the points where the hyperbola intersects its transverse axis. For the standard equation of a hyperbola \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] this axis is horizontal, and the vertices are given by the coordinates \((\pm a, 0)\). These points mark the closest points of the hyperbola to the center.

To find the vertices for a specific hyperbola, such as described by \( \frac{x^2}{4^2} - \frac{y^2}{3^2} = 1 \), you substitute the value of \(a\) (which in this case is 4) into the vertex formula. Thus, the vertices are at \((4, 0)\) and \((-4, 0)\). Remember:

• The transverse axis dictates the position of vertices: horizontal or vertical.
• Identifying \(a\) from the equation helps locate the vertices along this axis quickly.

Foci of Hyperbola

The foci of a hyperbola are two crucial points that lie along the transverse axis beyond the vertices. They hold the property that for any point on the hyperbola, the difference in distances to the foci is constant. These can be found for the equation form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), with the formula for the distance to the foci being\[ c^2 = a^2 + b^2 \] where \(c\) is the distance from the center to each focus along the transverse axis.

For the given hyperbola \(\frac{x^2}{4^2} - \frac{y^2}{3^2} = 1\), by substituting \(a = 4\) and \(b = 3\), you calculate \(c^2 = 16 + 9 = 25\), meaning \(c = \sqrt{25} = 5\). The foci will be at \((\pm 5, 0)\).

• The foci lie outside the vertices along the transverse axis.
• The value of \(c\) determines how far the foci are from the center on either side.

Equation of Hyperbola

A hyperbola's equation helps define its shape and orientation. The standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) zexpresses a hyperbola with a horizontal transverse axis. Here are a few key elements to understand:

• Transverse axis: This is the line that passes through the center, vertices, and foci of the hyperbola.
• Horizontal vs Vertical: If the \(x^2\) term precedes the \(y^2\) term, the transverse axis is horizontal. Conversely, if the \(y^2\) term is first, it's vertical.
• Components: Parameters \(a\) and \(b\) control the distances respective to vertices and constraints of the asymptotes.

Taking the sample equation \( \frac{x^2}{4^2} - \frac{y^2}{3^2} = 1 \), it can be observed that:

• The transverse axis is horizontal, indicated by the arrangement of terms.
• Values \(a = 4\) and \(b = 3\) assist in determining vertices and asymptotes details, thus sketching the hyperbola effectively.