Question

graph the hyperbola given by the following equation
\\(\frac{x^2}{16} - \frac{(y + 2)^2}{20} = 1\\)

Explanation:

Step1: Identify the standard form

The given equation is \(\frac{x^{2}}{16}-\frac{(y + 2)^{2}}{20}=1\), which is in the standard form of a hyperbola that opens horizontally: \(\frac{(x - h)^{2}}{a^{2}}-\frac{(y - k)^{2}}{b^{2}} = 1\), where \((h,k)\) is the center, \(a\) is the distance from the center to the vertices along the \(x\)-axis, and \(b\) is related to the conjugate axis. Here, \(h = 0\), \(k=-2\), \(a^{2}=16\) (so \(a = 4\)), and \(b^{2}=20\) (so \(b=\sqrt{20}=2\sqrt{5}\approx4.47\)).

Step2: Find the center

The center of the hyperbola is \((h,k)=(0,-2)\).

Step3: Determine the vertices

For a horizontal hyperbola, the vertices are at \((h\pm a,k)\). Substituting the values, we get \((0\pm4,-2)\), so the vertices are \((4,-2)\) and \((-4,-2)\).

Step4: Find the asymptotes

The equations of the asymptotes for a horizontal hyperbola are \(y - k=\pm\frac{b}{a}(x - h)\). Substituting \(h = 0\), \(k=-2\), \(a = 4\), and \(b = 2\sqrt{5}\), we get \(y+2=\pm\frac{2\sqrt{5}}{4}(x - 0)\), which simplifies to \(y+2=\pm\frac{\sqrt{5}}{2}x\) or \(y=\pm\frac{\sqrt{5}}{2}x-2\).

Step5: Sketch the hyperbola

1. Plot the center \((0,-2)\) on the coordinate plane.
2. Plot the vertices \((4,-2)\) and \((-4,-2)\).
3. Draw the rectangle centered at \((0,-2)\) with length \(2a = 8\) (along the \(x\)-axis) and width \(2b\approx8.94\) (along the \(y\)-axis). The corners of the rectangle are \((4, -2 + 2\sqrt{5})\), \((4,-2 - 2\sqrt{5})\), \((-4,-2 + 2\sqrt{5})\), and \((-4,-2 - 2\sqrt{5})\).
4. Draw the asymptotes through the corners of the rectangle (the lines \(y=\pm\frac{\sqrt{5}}{2}x-2\)).
5. Sketch the hyperbola opening to the left and right from the vertices, approaching the asymptotes as \(x\) moves away from the center.

Answer:

To graph the hyperbola \(\frac{x^{2}}{16}-\frac{(y + 2)^{2}}{20}=1\):

• Center: \((0, -2)\)
• Vertices: \((4, -2)\) and \((-4, -2)\)
• Asymptotes: \(y=\frac{\sqrt{5}}{2}x - 2\) and \(y=-\frac{\sqrt{5}}{2}x - 2\)

Plot the center, vertices, draw the asymptotes, and sketch the hyperbola opening horizontally (left and right) from the vertices, approaching the asymptotes.