Problem 54 - Precalculus: Mathematics for Calculus

Question

(a) For the hyperbola $$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$$ determine the values of \(a, b,\) and \(c,\) and find the coordinates of the foci \(F_{1}\) and \(F_{2}\). (b) Show that the point \(P\left(5, \frac{16}{3}\right)\) lies on this hyperbola. (c) Find \(d\left(P, F_{1}\right)\) and \(d\left(P, F_{2}\right)\). (d) Verify that the difference between \(d\left(P, F_{1}\right)\) and \(d\left(P, F_{2}\right)\) is \(2 a\).

Step-by-Step Solution

Answer

The coordinates of the foci are \(F_1(5,0)\) and \(F_2(-5,0)\). Point \(P\) lies on the hyperbola, and the difference in distances to the foci is 6, matching \(2a\).

Step 1: Identify Values of a, b, and Calculate c

For the hyperbola \(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\), recognize that this equation is of the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Here, \(a^2 = 9\) and \(b^2 = 16\). So, \(a = 3\) and \(b = 4\). To find \(c\), use the equation \(c^2 = a^2 + b^2\): \(c^2 = 9 + 16 = 25\) which gives \(c = 5\).

Step 2: Determine Coordinates of the Foci

The foci \(F_1\) and \(F_2\) are located at \((c, 0)\) and \((-c, 0)\) for a hyperbola with a horizontal transverse axis. Given \(c = 5\), the coordinates are \(F_1(5, 0)\) and \(F_2(-5, 0)\).

Step 3: Check if Point P Lies on the Hyperbola

For point \(P(5, \frac{16}{3})\), substitute into the hyperbola equation: \(\frac{5^2}{9} - \left(\frac{16}{3}\right)^2/16 = 1\). Calculate: \(\frac{25}{9} - \frac{256}{144} = \frac{256}{144}\). Simplifying: \(\frac{25}{9} - \frac{64}{36} = 1\), which verifies the calculation as \(\frac{100}{36} - \frac{64}{36} = 1\). Thus, point \(P\) satisfies the hyperbola equation.

Step 4: Calculate Distance d(P, F1)

The distance formula is \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). For \(d(P, F_1)\), compute \(\sqrt{(5 - 5)^2 + \left(\frac{16}{3} - 0\right)^2}\), which simplifies to \(\frac{16}{3}\).

Step 5: Calculate Distance d(P, F2)

For \(d(P, F_2)\): Use \(\sqrt{(5 - (-5))^2 + \left(\frac{16}{3} - 0\right)^2}\). This becomes \(\sqrt{(10)^2 + \left(\frac{16}{3}\right)^2}\), which simplifies to \(\sqrt{100 + \frac{256}{9}} = \sqrt{\frac{1156}{9}} = \frac{34}{3}\).

Step 6: Verify the Difference is 2a

The difference \(d(P, F_2) - d(P, F_1)\) is \(\frac{34}{3} - \frac{16}{3} = \frac{18}{3} = 6\). Since \(2a = 2 \times 3 = 6\), the condition is verified.

Key Concepts

Foci

In the context of hyperbolas, the foci (plural of focus) are two fixed points used to define the curve. For a hyperbola with a horizontal transverse axis, like the one described in the equation \(\frac{x^2}{9} - \frac{y^2}{16} = 1\), the foci are found along the x-axis.

To determine their placement, you start by calculating the distance \(c\) from the center to each focus. This is done using the relation \(c^2 = a^2 + b^2\). Given that \(a = 3\) and \(b = 4\), we calculate \(c = \sqrt{25} = 5\).

Thus, the foci \(F_1\) and \(F_2\) are at coordinates \((c, 0)\) and \((-c, 0)\), or more specifically, at \((5, 0)\) and \((-5, 0)\). These points are fundamental as they help to maintain the hyperbola's unique property: the difference in distances from any point on the hyperbola to each focus remains constant.

Distance Formula

The distance formula provides a straightforward method for calculating the distance between two points in a plane. For two points, \((x_1, y_1)\) and \((x_2, y_2)\), the formula is given by:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

In our exercise, we use this formula to determine the distances \(d(P, F_1)\) and \(d(P, F_2)\), where \(P\) is the point \((5, \frac{16}{3})\) and the foci \(F_1\) and \(F_2\) have coordinates \((5,0)\) and \((-5,0)\) respectively.

Specifically:

• \(d(P, F_1) = \sqrt{(5 - 5)^2 + \left(\frac{16}{3} - 0\right)^2} = \frac{16}{3}\)
• \(d(P, F_2) = \sqrt{(5 - (-5))^2 + \left(\frac{16}{3} - 0\right)^2} = \frac{34}{3}\)

These computations are critical for verifying that the differences in distances reflect the constant characteristic of the hyperbola.

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. Depending on the angle of the intersecting plane, various types of conic sections are formed, including circles, ellipses, parabolas, and hyperbolas.

A hyperbola, like the exercise at hand, results when the plane intersects both nappes of the cone but not at the vertex. Hyperbolas are characterized by their pair of symmetric open curves. The defining property of a hyperbola is that for any point on the hyperbola, the absolute difference of the distances to the two foci is constant. This is unlike other conics such as ellipses where the sum is constant.

Hyperbolas can be further classified by whether the transverse axis is horizontal or vertical. In this example, the transverse axis is horizontal, as evidenced by the equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Understanding these properties is essential for graphing hyperbolas and solving complex geometric problems.

Transverse Axis

The transverse axis is a key concept when analyzing hyperbolas. This axis is the line segment that passes through the center of the hyperbola and its foci. Its length is directly related to the parameter \(a\) of the hyperbola's standard form equation.

In the equation \(\frac{x^2}{9} - \frac{y^2}{16} = 1\), the transverse axis lies horizontally along the x-axis. This is because the term associated with \(x^2\) comes first, indicating a horizontal orientation. Since \(a^2 = 9\), we have \(a = 3\) and thus the transverse axis has a length of \(2a = 6\).

The transverse axis plays a vital role in the shape and orientation of the hyperbola, affecting where its vertices and foci lie. Being able to identify and correctly interpret the transverse axis can simplify the study of hyperbolas and aid in solving related mathematical problems.