Problem 46 - 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 foci are at (5, 0) and (-5, 0). The point P lies on the hyperbola. Distances are 16/3 and 34/3. The difference is 6.

Step 1: Identify Equation Values

The given equation of the hyperbola is \(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\). By comparing this with the standard hyperbola equation \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can immediately identify that \(a^2 = 9\) and \(b^2 = 16\). Thus, \(a = 3\) and \(b = 4\).

Step 2: Calculate c

For hyperbolas, \(c\) is found using the relationship \(c^2 = a^2 + b^2\). Substituting \(a^2 = 9\) and \(b^2 = 16\), we get \(c^2 = 9 + 16 = 25\). Therefore, \(c = 5\).

Step 3: Determine Foci Coordinates

The foci \(F_1\) and \(F_2\) for hyperbolas with a transverse axis in the x-direction are located at \((\pm c, 0)\). Here, since \(c = 5\), the foci are at \((5, 0)\) and \((-5, 0)\).

Step 4: Check if Point Lies on Hyperbola

Substitute the point \(P\left(5, \frac{16}{3}\right)\) into the hyperbola equation \(\frac{x^2}{9} - \frac{y^2}{16} = 1\). This gives \(\frac{5^2}{9} - \left(\frac{\left(\frac{16}{3}\right)^2}{16}\right)\). Simplifying, we get \(\frac{25}{9} - \frac{256}{144} = \frac{25}{9} - \frac{64}{36} = \frac{25 \times 4}{36} - \frac{64}{36} = 1\). Thus, the point lies on the hyperbola.

Step 5: Calculate Distances from Foci

The distance \(d(P, F_1)\) is calculated using the distance formula between \(P(5, \frac{16}{3})\) and \(F_1(5, 0)\). This is \(d(P, F_1) = \sqrt{(5-5)^2 + (\frac{16}{3}-0)^2} = \frac{16}{3}\). The distance \(d(P, F_2)\) is \(d(P, F_2) = \sqrt{(5+5)^2 + (\frac{16}{3}-0)^2} = \sqrt{100 + \left(\frac{16}{3}\right)^2} = \sqrt{100 + \frac{256}{9}}\). Further simplifying gives \(\sqrt{100 + \frac{256}{9}} = \sqrt{900/9 + 256/9} = \sqrt{1156/9} = \frac{34}{3}\).

Step 6: Verify Difference

For hyperbolas, the formula \(|d(P, F_1) - d(P, F_2)| = 2a\) should hold, where \(2a = 6\). Calculating the difference, \(|\frac{16}{3} - \frac{34}{3}| = |0 - 5| = 6\), meets the condition \(2a = 6\). Hence, the relationship is verified.

Key Concepts

Conic Sections

Conic sections are curves obtained by intersecting a plane with a double-napped cone. These intersections result in different shapes including circles, ellipses, parabolas, and hyperbolas. Each type of conic section has unique properties and equations that define their shapes.
Hyperbolas, specifically, are characterized by two separate curves that mirror each other. These arise from cutting a cone parallel to its axis but not at that axis itself. Understanding hyperbolas and other conic sections is crucial as they frequently appear in various fields including physics, engineering, and astronomy.

Foci of Hyperbola

The foci of a hyperbola are two fixed points located along the transverse axis, which is the axis that passes through the center of the hyperbola and its vertex points. These foci are important because every point on the hyperbola has a constant difference in distances to these two points.
To find the foci of a hyperbola in the standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), we can use the equation \(c^2 = a^2 + b^2\). The coordinates of the foci \((\pm c, 0)\) indicate that for a hyperbola oriented along the x-axis, the foci are located symmetrically on either side of the center, usually at \( (\pm c, 0)\). Hence, understanding foci helps in sketching and analyzing the hyperbola's geometry.

Distance Formula

The distance formula is a fundamental tool in geometry for calculating the distance between two points in a Cartesian plane. The formula is given as \( \text{Distance} = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \). This calculation is critical in determining how far one point in space is from another.
In the hyperbola problem, distances from the point \(P\) to the foci \(F_1\) and \(F_2\) are calculated using this formula, showing the unique property of hyperbolas where these distances have a constant difference. This illustrates how the general distance formula has specific applications for conic sections like hyperbolas.

Verifying Points on a Conic

To verify if a point lies on a conic section, we substitute the coordinates of the point into the conic's equation. If the equation holds true, the point is on the conic.
For hyperbolas given by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), place the coordinates of the point into the equation. If both sides equate, the point is confirmed to be on the hyperbola. This step is vital as it validates our understanding of conic equations and their graphical representations, ensuring the point abides by the hyperbola's defined path.

Standard Form of Hyperbola Equation

The standard form of a hyperbola equation is central to understanding its properties and graphing it correctly. It generally is written as \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) for hyperbolas oriented along the x-axis, and \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\) for those along the y-axis.
This format tells us specific measures such as the semi-major axis length \(a\) and semi-minor axis length \(b\), which are pivotal for finding other characteristics, such as the foci. Understanding the standard form helps in recognizing and solving problems related to hyperbolas, allowing for accurate calculations of properties like the transverse and conjugate axes, and confirming the geomorphic nature of the curve.