Problem 10 - Algebra and Trigonometry

Question

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±√2, -6); Asymptotes: y=x-14, y=-x+2.

Step 1: Identify the Standard Form

The given equation is \((x-8)^2 - (y+6)^2 = 1\). This is in the form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), indicating that the hyperbola is oriented horizontally.

Step 2: Determine the Center

In the standard form \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), the center is \((h, k)\). By comparing, we have \(h = 8\) and \(k = -6\). Therefore, the center is \((8, -6)\).

Step 3: Find the Vertices

The vertices of a horizontally oriented hyperbola are \((h \pm a, k)\). Since we have it in the form \((x-8)^2 - (y+6)^2 = 1\), we identify \(a^2 = 1\), so \(a = 1\). This gives the vertices as \((8 \pm 1, -6) = (9, -6)\) and \((7, -6)\).

Step 4: Calculate the Foci

For a horizontally oriented hyperbola, the foci are \((h \pm c, k)\), where \(c = \sqrt{a^2 + b^2}\). Here, since \(b^2\) is absent, we consider it as 1, thus implying \(c = \sqrt{1+1} = \sqrt{2}\). Therefore, the foci are \((8 \pm \sqrt{2}, -6)\).

Step 5: Determine the Asymptotes

In a horizontally oriented hyperbola \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\), the asymptotes have the equations \(y - k = \pm \frac{b}{a}(x-h)\). With our values, it becomes \(b/a = 1\), so the asymptotes are \(y + 6 = \pm (x - 8)\) which simplifies to \(y = x - 14\) and \(y = -x + 2\).

Step 6: Sketch the Graph

Draw the axes and plot the center at \((8, -6)\). Place the vertices at \((7, -6)\) and \((9, -6)\), then the foci very close to these points, at approximately \((8+\sqrt{2}, -6)\) and \((8-\sqrt{2}, -6)\). Use the asymptotes \(y = x-14\) and \(y = -x+2\) as guides to draw the branches of the hyperbola opening left and right.

Key Concepts

Center of Hyperbola

The center of a hyperbola is the midpoint between its two branches and serves as a reference point for graphing the structure. To find the center, we look at the standard form equation of the hyperbola. The equation given in this exercise is \[(x-8)^2 - (y+6)^2 = 1,\]which corresponds to the standard form \[(x-h)^2/a^2 - (y-k)^2/b^2 = 1.\]By comparing the two, we identify that the center \((h, k)\)is at \((8, -6)\).

• "\(h\)" is the number subtracted from "\(x\)" (which is 8 here)
• "\(k\)" is the number subtracted from "\(y\)" (considering the sign change, here it becomes -6)

Understanding this concept will make it much easier to start sketching the hyperbola. Placing the center correctly on the graph is the first step towards a correct illustration.

Vertices of Hyperbola

The vertices of a hyperbola are crucial for defining the extent of its branches. They lie along the transverse axis, which is the real axis for the hyperbola, running through its center.For a horizontally oriented hyperbola, the vertices are calculated by adjusting "\(h\)" with "\(a\)". The equation for the vertices is \((h \pm a, k)\).In our example, we identified:- "\(a^2 = 1\)", thus "\(a = 1\)"- Center is \((8, -6)\)To calculate the vertices:- Place the first at \((8+1, -6) = (9, -6)\)- The second at \((8-1, -6) = (7, -6)\)

• Vertices provide the width of the hyperbola and indicate where the graph opens.
• They are positioned symmetrically around the center on the transverse axis.

Plotting these on the graph helps define the curve's overall structure and bounds.

Foci of Hyperbola

The foci of a hyperbola are points located along the transverse axis within each branch but outside of its vertices. They are essential because they help define the shape of the hyperbola.For a horizontally oriented hyperbola, the formula to find the foci is \((h \pm c, k)\),where \(c = \sqrt{a^2 + b^2}\).From the equation \((x-8)^2 - (y+6)^2 = 1\)We calculate:- "\(a^2 = 1\)", actual "\(b^2 = 1\)" (default since it is absent), which gives "\(c = \sqrt{1 + 1} = \sqrt{2}\)"Thus, the points are:- First focus at \((8+\sqrt{2}, -6)\)- Second focus at \((8-\sqrt{2}, -6)\)

• Foci are inside each curve branch, lying closer than the vertices.
• They drive how the hyperbola bends and sharpens towards these points.

Marking the foci helps deepen your understanding of how the hyperbola will continue bending within and beyond the graph's sketched image.

Asymptotes of Hyperbola

The asymptotes of a hyperbola are straight lines that act as invisible boundaries, which the branches of the hyperbola will approach but never touch. Asymptotes guide the curve of the hyperbola and suggest its spread.For a horizontally oriented hyperbola, the equations of its asymptotes are typically\[y - k = \pm \frac{b}{a}(x - h).\]In our example:- "\(b/a = 1\)" (since "\(b^2 = a^2 = 1\)"- Center is at \((h, k) = (8, -6)\)Substitute to find:- First asymptote equation: \(y + 6 = (x - 8)\) which simplifies to \(y = x - 14\)- Second asymptote equation: \(y + 6 = -(x - 8)\) which simplifies to \(y = -x + 2\)

• Asymptotes cross at the center point and diverge outward from the hyperbola's branches.
• They set the ultimate path of the hyperbola's curve as it extends indefinitely.

Drawing these helps in visualizing how the hyperbola behaves as it moves away from its center, ensuring accurate graph representation of the equation.