Problem 45 - Calculus Early Transcendentals

Question

Find an equation for the conic that satisfies the given conditions. $$\begin{array}{l}{\text { Hyperbola, vertices }(-3,-4),(-3,6)} \\ {\text { foci }(-3,-7),(-3,9)}\end{array}$$

Step-by-Step Solution

Answer

The equation of the hyperbola is \( \frac{(y - 1)^2}{25} - \frac{(x + 3)^2}{39} = 1 \).

Step 1: Determine the Orientation

Since the vertices and foci have the same x-coordinate, the hyperbola is vertical. Thus, its standard equation will be of the form: \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \]

Step 2: Find the Center

The center \((h, k)\) is the midpoint of the line segment connecting the vertices. The midpoint of \((-3, -4)\) and \((-3, 6)\) is \((-3, \frac{-4 + 6}{2}) = (-3, 1)\). So, the center is \((h, k) = (-3, 1)\).

Step 3: Calculate \(a\)

The distance between the vertices is 10. Since the form is vertical, the distance between the vertices is \(2a\). Therefore, \(2a = 10\) implies \(a = 5\).

Step 4: Calculate \(c\)

The distance between the foci \((-3, -7)\) and \((-3, 9)\) is 16. Since the form is vertical, the distance between the foci is \(2c\). Therefore, \(2c = 16\) implies \(c = 8\).

Step 5: Use the Relationship \(c^2 = a^2 + b^2\)

Using the relationship \(c^2 = a^2 + b^2\), we substitute \(c = 8\) and \(a = 5\):\[8^2 = 5^2 + b^2\]\[64 = 25 + b^2\]\[b^2 = 64 - 25\]\[b^2 = 39\].

Step 6: Write the Equation of the Hyperbola

Substitute \(h = -3\), \(k = 1\), \(a^2 = 25\), and \(b^2 = 39\) into the standard form for a vertical hyperbola:\[ \frac{(y - 1)^2}{25} - \frac{(x + 3)^2}{39} = 1 \].

Key Concepts

Vertices

Vertices are crucial points for understanding the shape of a hyperbola. In the context of our hyperbola, the vertices are given as

• (-3, -4)
• (-3, 6)

These vertices have the same x-coordinate, indicating that the hyperbola is oriented vertically, extending above and below a central point along the y-axis.
The vertices mark the turning points of the hyperbola's two branches and help us determine the major axis, which in this case is vertical.
For a hyperbola, the distance between the vertices is key for further calculations and is noted as \(2a\).
If you were to visualize the hyperbola, these vertices would be the closest points on each of the two branches, located on the line that bisects and is perpendicular to the transverse axis of the hyperbola.

Foci

Foci (singular: focus) of a hyperbola are points that help define the shape's geometry more precisely.
Our hyperbola's foci are located at

• (-3, -7)
• (-3, 9)

These too share a common x-coordinate, reinforcing the vertical orientation.
The foci are positioned further away from the center than the vertices, lying along the same transverse axis.
The distance between the foci is equal to \(2c\), a measure when split into two is referred to as 'c'.
The relationship between the distances from any point on the hyperbola to the two foci is constant, which is the property that extends the open branches in opposite directions.
This constant uses in defining the hyperbola's characteristic equation, contributing critical information for solving hyperbolic coverage problems.

Equation of a Conic

When considering a hyperbola, its standard mathematical representation is known as the equation of a conic.
For vertical hyperbolas like ours, the standard form of the equation is given by:\[\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\]This formula involves the squared terms of its variables, with numerators concerned with their respective deviations from the center point \(h, k\).
The denominators \(a^2\) and \(b^2\) determine the distances resulting from the vertices and foci relationships.
In practical terms:

• "a" is derived from half the distance between the vertices,
• "c" is half the distance between the foci, and buoyed by the relationship \(c^2 = a^2 + b^2\),
• \"b\" can be computed as \(b^2 = c^2 - a^2\).

This systematic approach allows us to obtain the unique equation of a hyperbola that matches the given conditions; visible here as:\[\frac{(y - 1)^2}{25} - \frac{(x + 3)^2}{39} = 1\]Such equations help model real-world phenomena, ranging from satellite orbits to specific acoustics scenarios.