91Ó°ÊÓ

One of the fundamental aspects of understanding hyperbolas is recognizing their equation in standard form. The standard form for a hyperbola varies slightly depending on its orientation.

For a vertical hyperbola, the equation is \[\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\]This is different from a horizontal hyperbola, which is expressed as:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]The notable feature here is the position of the negative sign, which dictates the orientation. In our specific problem, the equation is \(\frac{(y+1)^2}{16} - \frac{(x-4)^2}{36} = 1\). This tells us it's a vertical hyperbola, as the squared \(y\)-term comes first. Identifying the standard form is crucial as it aids in determining other critical components of the hyperbola.

Understanding the center of a hyperbola helps anchor other components like vertices and foci. The center of a hyperbola in its standard form \(\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1\) is given by the point \((h, k)\).

In our example equation, \(\frac{(y+1)^2}{16} - \frac{(x-4)^2}{36} = 1\), we see:

• \(h = 4\)
• \(k = -1\)

Thus, the center of this hyperbola is \((4, -1)\).This central point acts as the reference point from which distances to vertices and foci are calculated. It serves as the midpoint of the line segment connecting its foci.

Vertices play a key role in defining the shape and extent of a hyperbola. For a vertical hyperbola, the vertices are located at \((h, k \pm a)\).

To find \(a\), we typically look at the denominator of the squared \(y\)-term, where \(a^2 = 16\). Hence, \(a = 4\).

• The vertices thus are at \((4, -1 + 4)\) and \((4, -1 - 4)\).
• Simplifying gives the vertices at \((4, 3)\) and \((4, -5)\).

These vertices help define the spread of the hyperbola along the vertical axis. By plotting them, the initial sketch of the hyperbola’s central shape is established.

Foci extend further out than the vertices and help further define the hyperbola's shape and orientation. For a hyperbola, the foci are calculated using the formula \((h, k \pm c)\), where \(c^2 = a^2 + b^2\).

In this equation:

• \(b^2\) is given as 36, thus \(b = 6\).
• So, \(c^2 = 16 + 36 = 52\), leading to \(c = \sqrt{52} = 2\sqrt{13}\).

The precise location of the foci, therefore, is:

• \((4, -1 + 2\sqrt{13})\)
• \((4, -1 - 2\sqrt{13})\)

These points are essential for plotting an accurate hyperbola, showcasing how the shape extends infinitely along a particular trajectory from these focal points.