Problem 7 - Algebra and Trigonometry

Question

find the standard form of the equation of each hyperbola satisfying the given conditions. $$ \text { Foci: }(-4,0),(4,0) ; \text { vertices: }(-3,0),(3,0) $$

Step-by-Step Solution

Answer

The standard form of the equation of the hyperbola is \(\frac{x^2}{9} - \frac{y^2}{7} = 1\)

Step 1: Find the center of the hyperbola

In this case, we can see that the center of the hyperbola is halfway between either pair of vertices or foci. Thus, since the vertices are at the points (-3,0) and (3,0), the center of the hyperbola is at the origin (0,0).

Step 2: Identify a and c

Since the vertices denote the length of the semi-major axis on the hyperbola, we know that \(a\) equals 3. Using the same rationale, the foci denote the distance from the center to the foci, \(c\). Here, \(c\) equals 4.

Step 3: Calculate for b

For a hyperbola, we have the relationship \(c^2 = a^2 + b^2\). We can interpret these values as follows: \(a\) is the distance from the center to a vertex, \(b\) is the distance from the center to a co-vertex, and \(c\) is the distance from the center to a focus. We can rearrange the equation to solve for \(b^2\): \(b^2 = c^2 - a^2 = 16 - 9 = 7\).

Step 4: Formulate the equation of the hyperbola

The standard form for a horizontally oriented hyperbola centered at the origin (h,k) is given by \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Plugging our values in, we get \(\frac{x^2}{9} - \frac{y^2}{7} = 1\)

Key Concepts

Center of Hyperbola

The center of a hyperbola is like its middle point, similar to the center of a circle. This point is pivotal because it helps us determine the way the hyperbola spreads out in the coordinate plane. To find this point, you need to look for the midpoint between either of the pairs of foci or vertices. They are usually given in the problem. For instance, if the vertices are at

• (-3,0) and (3,0),

we find the midpoint to determine the center. The math behind finding a midpoint is straightforward; you average the x-coordinates and y-coordinates separately. Here, the midpoint between the given vertices is the origin (0,0). Knowing this center point helps set the stage for crafting the equation of the hyperbola.

Semi-Major Axis

The term 'semi-major axis' refers to the axis along which the hyperbola stretches the farthest from its center. It's similar to the main stretch of an ellipse, but since an ellipse doesn't open up like a hyperbola, the terms apply slightly differently.In a hyperbola, the vertices reveal the semi-major axis length. They show us how the hyperbola reaches out from its center. By determining the distance from the center to a vertex, you get the length of the semi-major axis. This length is labeled as

• \(a\)
.For example, between vertices
• (-3,0) and (3,0),
we measure the span from the center (0,0) to either vertex, which is 3 units. Hence, for this hyperbola, \(a = 3\). This measurement helps form the equation structure for the hyperbola, offering clarity on its spatial orientation.

Hyperbola Equation

A hyperbola's equation ties all its traits together into a neat algebraic package. For a hyperbola centered at the origin and extending sideways, its standard form equation looks like this:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]Here, \((h,k)\) denotes the center, while \(a\) and \(b\) stand for the semi-major and semi-minor axes' lengths, respectively. By inputting the determined values for our specific hyperbola, where the center is (0,0), \(a = 3\), and we've figured \(b^2 = 7\), the equation becomes:\[\frac{x^2}{9} - \frac{y^2}{7} = 1\]This equation encapsulates the hyperbola's form. The value of \(a\) came from the distance to the vertices, and \(b\) was derived from rearranging the relationship formula \(c^2 = a^2 + b^2\) postulating the hyperbola's structure. The equation paints a full picture of the hyperbola's stretch and symmetry, offering a formulaic insight into its geometric identity.