Problem 38 - Precalculus with Limits

Question

In Exercises 35-46, find the standard form of the equation of the hyperbola with the given characteristics. Vertices: \((-2, 1), (2, 1); \quad\) foci: \((-3, 1), (3, 1)\)

Step-by-Step Solution

Answer

The standard form of the hyperbola with vertices at \((-2, 1)\) and \((2, 1)\), and foci at \((-3, 1)\) and \((3, 1)\) is \(\frac{x^{2}}{4}-\frac{(y-1)^{2}}{5}=1\).

Step 1: Define Given Points

Our vertices are at points \((-2, 1)\) and \((2, 1)\), and foci are at \((-3, 1)\) and \((3, 1)\). The x-coordinates of vertices and foci indicate that the hyperbola is horizontally oriented (opens to the left and right). The center of the hyperbola will be halfway between the vertices or the foci.

Step 2: Locate the Center

The center of hyperbola can be found by calculating the midpoint between the vertices (or the foci). The midpoint is calculated by averaging the x-coordinates and the y-coordinates of the vertices (or foci). So, the x-coordinate of the center is \(\frac{-2+2}{2}=0\) and the y-coordinate of the centre is \(\frac{1+1}{2}=1\). Hence, the center is at \((0, 1)\).

Step 3: Calculate the Lengths of Semi-Major and Semi-Minor Axes

The distance from the center to a vertex, gives us the value of 'a', which is the semi-major axis. Since the vertices are at points \((-2, 1)\) and \((2, 1)\), 'a' would be equal to \(2 - 0 = 2\).\nThe distance from the center to a focus, will give us the value of 'c', which is the distance from the center to either focus. Considering our foci are at points \((-3, 1)\) and \((3, 1)\), 'c' would be equal to \(3 - 0 = 3\).\nWith 'a' and 'c', we can determine the semi-minor axis, 'b', using the relationship \(c^{2}=a^{2}+b^{2}\). Solving for 'b' gives us \(b=\sqrt{c^{2}-a^{2}}=\sqrt{9-4}=\sqrt{5}\).

Step 4: Write the Standard Form Hyperbola Equation

Finally, the standard form of the hyperbola is \(\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1\) for a horizontally oriented hyperbola. Here, (h,k) is the center of hyperbola. Plugging in our known values, we obtain the equation \(\frac{(x-0)^{2}}{2^{2}}-\frac{(y-1)^{2}}{(\sqrt{5})^{2}}=1\), which simplifies to \(\frac{x^{2}}{4}-\frac{(y-1)^{2}}{5}=1\). This is the standard equation of the hyperbola.

Key Concepts

Standard Form of Hyperbola

Understanding the standard form of a hyperbola is crucial in identifying its characteristics and plotting it properly on a coordinate plane. A hyperbola has two types of standard forms depending on its orientation — horizontal or vertical. For a horizontally oriented hyperbola, the standard form is:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]Similarly, for a vertically oriented hyperbola, it is:\[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]Where:

• \((h, k)\) is the center of the hyperbola.
• 'a' represents the semi-major axis length.
• 'b' represents the semi-minor axis length.

By plugging in the correct values based on the problem, the proper equation of the hyperbola can be written.

Vertices and Foci

Vertices and foci are key components that dictate the shape and properties of a hyperbola. These points lie along the major axis of the hyperbola. This axis, being either horizontal or vertical, helps determine the form of the hyperbola’s equation.

Understanding the Role of Vertices

Vertices are located at the ends of the semi-major axis and define the width of the hyperbola. In this problem, the vertices are located at \((-2, 1)\) and \((2, 1)\). The midpoint of these vertices provides the hyperbola's center.

Importance of Foci

Foci lie inside each curve of the hyperbola and are equidistant from the center. They play a pivotal role in determining the eccentricity of the hyperbola. In our case, the foci are at \((-3, 1)\) and \((3, 1)\). Notably, the foci being at positions further from the center than the vertices confirm the extension of the hyperbola.

Equation of the Hyperbola

Once the center, vertices, and foci are identified, crafting the equation of the hyperbola becomes straightforward. The hyperbola's equation follows a standard formula, as detailed previously.To derive the specific equation for the given problem:

• The center of the hyperbola \((h, k)\) is found at \((0, 1)\).
• The distance from the center to the vertices provides 'a', the semi-major axis, which is 2, calculated as the distance from 0 to 2.
• The foci provide 'c', calculated as 3, the distance from 0 to 3.

Substituting these values into the standard form yields the equation:\[ \frac{(x-0)^2}{2^2} - \frac{(y-1)^2}{(\sqrt{5})^2} = 1 \]This simplifies to the equation:\[ \frac{x^2}{4} - \frac{(y-1)^2}{5} = 1 \]This particular hyperbolic equation is a direct expression of its geometric properties.

Semi-Major and Semi-Minor Axes Calculation

Understanding the calculation of the semi-major and semi-minor axes is essential in characterizing a hyperbola. The semi-major axis, often denoted by 'a', is the larger of the distances from the center to a vertex.

Calculation of Semi-Major Axis

In this problem, the distance known as 'a' is calculated from the center \((0, 1)\) to the vertex \((2, 1)\). This distance is straightforwardly calculated to be 2 units.

Determining the Semi-Minor Axis

The semi-minor axis, 'b', can be determined using the relationship between 'a', 'b', and 'c' (distance to the foci):\[ c^2 = a^2 + b^2 \]Given that 'c' equals 3 and 'a' equals 2, 'b' can be solved as:\[ b = \sqrt{c^2 - a^2} = \sqrt{9 - 4} = \sqrt{5} \]This calculation ensures that all paths from the center to the various defining points of the hyperbola are accurately represented, allowing us to fully understand its dimensions.