Problem 14 - Algebra and Trigonometry

Question

\(13-16\) . Find the center, foci, vertices, and asymptotes of the hyperbola. Then sketch the graph. $$ (x-8)^{2}-(y+6)^{2}=1 $$

Step-by-Step Solution

Answer

Center: \((8, -6)\), Vertices: \((9, -6)\), \((7, -6)\), Foci: \((8 \pm \sqrt{2}, -6)\), Asymptotes: \(y = x - 14\), \(y = -x + 2\).

Step 1: Identify the Hyperbola Form

The given equation is \((x-8)^2 - (y+6)^2 = 1\), which is already in the standard form \((x-h)^2 - (y-k)^2 = 1\). This represents a hyperbola centered at \((h, k)\) with horizontal transverse axis.

Step 2: Find the Center

The center \((h, k)\) is found directly from the equation form. Here, \(h = 8\) and \(k = -6\). So, the center is \((8, -6)\).

Step 3: Identify and Calculate the Vertices

For a hyperbola in the form \((x-h)^2 - (y-k)^2 = 1\), the vertices are \((h \pm a, k)\). Since there is no coefficient under the \((x-h)^2\) term, \(a = 1\). The vertices are thus \((8 \pm 1, -6)\), or \((9, -6)\) and \((7, -6)\).

Step 4: Find the Foci

The distance from the center to each focus is \(c\), where \(c = \sqrt{a^2 + b^2}\). Here, \(a = 1\) and \(b = 1\) (since the coefficient under \((y+6)^2\) is \(-1\)). Hence, \(c = \sqrt{1^2 + 1^2} = \sqrt{2}\). The foci are thus \((8 \pm \sqrt{2}, -6)\).

Step 5: Determine the Asymptotes

The asymptotes of the hyperbola are given by the equations \(y - k = \pm \frac{b}{a}(x - h)\). Plugging in \(a = 1\), \(b = 1\), \(h = 8\), and \(k = -6\), the asymptotes are \(y + 6 = \pm 1(x - 8)\), or simplified: \(y = x - 14\) and \(y = -x + 2\).

Step 6: Sketch the Hyperbola

To sketch the hyperbola, plot the center at \((8, -6)\), plot the vertices at \((9, -6)\) and \((7, -6)\), and draw the asymptotes using the equations \(y = x - 14\) and \(y = -x + 2\). Draw the hyperbola opening horizontally along these guides.

Key Concepts

Center of Hyperbola

The center of a hyperbola is a key component when defining its shape and structure in a coordinate plane. For the hyperbola described by the equation \((x-8)^2 - (y+6)^2 = 1\), we can identify the center easily from the equation's standard form \((x-h)^2 - (y-k)^2 = 1\). Here, \(h\) and \(k\) are coordinates that give us the center point. In this case, we have \(h = 8\) and \(k = -6\). Thus, the center of the hyperbola is at the point \((8, -6)\). This point is important because it acts as a reference for determining other aspects such as vertices, foci, and the sketching of the graph.
The center divides the hyperbola into symmetrical sections, providing balance and structure to the shape it makes on the graph.

Vertices of Hyperbola

Vertices are significant points on a hyperbola that lie on its transverse axis. For our given hyperbola, the transverse axis is horizontal. The formula \((x-h)^2 - (y-k)^2 = 1\) also helps us identify where the vertices land. If the equation is in this form, the vertices can be found at \((h \pm a, k)\).
In the equation \((x-8)^2 - (y+6)^2 = 1\), the value of \(a\) is \(1\) since there is no apparent coefficient (which implies it is \(1\)). Thus, we calculate the vertices to be at \((8 \pm 1, -6)\).

• This results in the points \((9, -6)\)
• and \((7, -6)\)

These are the points where the shape of the hyperbola is closest to the center in the horizontal direction.

Foci of Hyperbola

The foci of a hyperbola are special points located along the transverse axis, further away from the center than the vertices. They play a crucial role in defining the shape of the hyperbola. The separation between the center and these points is greater than that of the vertices, highlighting the hyperbola's distinct nature.
To compute the foci for the equation \((x-8)^2 - (y+6)^2 = 1\), we first need the value of \(c\), where \(c\) is the distance from the center to each focus. The formula to find \(c\) is: \[ c = \sqrt{a^2 + b^2} \] With \(a = 1\) and \(b = 1\), this gives us:\[ c = \sqrt{1^2 + 1^2} = \sqrt{2} \] Therefore, the foci are located at \((8 \pm \sqrt{2}, -6)\).
These points help in the sketching phase, ensuring the hyperbola's paths curve appropriately away from the center.

Asymptotes of Hyperbola

The asymptotes of a hyperbola are straight lines that the hyperbola approaches but never actually touches as it extends indefinitely in either direction. These lines are critical as they guide the overall direction and spacing of the hyperbola's curves. Asymptotes almost act like invisible boundaries for the hyperbola's wings.
For the equation \((x-8)^2 - (y+6)^2 = 1\), the asymptotes can be calculated using the formula:\[ y - k = \pm \frac{b}{a}(x - h) \] Plugging in the values \(a = 1\), \(b = 1\), \(h = 8\), and \(k = -6\), we find:\[ y + 6 = \pm 1(x - 8) \] which simplifies to:

• \(y = x - 14\)
• and \(y = -x + 2\)

These equations provide the paths that guide the arms of the hyperbola, framing the sketch and helping with plotting the graph accurately.