Problem 9 - Algebra and Trigonometry

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$ \frac{x^{2}}{4}-\frac{y^{2}}{16}=1 $$

Step-by-Step Solution

Answer

Vertices: (2, 0), (-2, 0); Foci: (2\sqrt{5}, 0), (-2\sqrt{5}, 0); Asymptotes: y=2x, y=-2x.

Step 1: Identify the Standard Form

The equation provided is \( \frac{x^{2}}{4} - \frac{y^{2}}{16} = 1 \). This equation resembles the standard form of a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \( a^2 = 4 \) and \( b^2 = 16 \). Therefore, \( a = 2 \) and \( b = 4 \).

Step 2: Find the Vertices

For a hyperbola of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are located at \(( \pm a, 0)\). Substituting \( a = 2 \), the vertices are \((2, 0)\) and \((-2, 0)\).

Step 3: Calculate the Foci

The foci are given by \(( \pm c, 0)\) where \( c = \sqrt{a^2 + b^2} \). Here, \( c = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \). Thus, the foci are \((2 \sqrt{5}, 0)\) and \((-2 \sqrt{5}, 0)\).

Step 4: Determine the Asymptotes

The equations for the asymptotes of this hyperbola are given by \( y = \pm \frac{b}{a}x \). Hence, \( y = \pm \frac{4}{2}x \), which simplifies to \( y = \pm 2x \).

Step 5: Sketch the Graph

To sketch the hyperbola, plot the vertices at \((2, 0)\) and \((-2, 0)\). Draw the asymptotes \( y = 2x \) and \( y = -2x \), crossing through the center at \((0,0)\). The hyperbola will approach but never intersect these asymptotes, opening left and right from the vertices.

Key Concepts

Vertices of a Hyperbola

Vertices are crucial components of a hyperbola as they are points where the hyperbola intersects its transverse axis. They essentially mark the closest distance of the hyperbola to the center of the graph.

For the equation of our hyperbola, \( \frac{x^2}{4} - \frac{y^2}{16} = 1 \), the standard form, \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), helps us determine the vertices. Here, \( a^2 = 4 \), giving us \( a = 2 \).

Therefore, the vertices for this equation are determined by the coordinates

• \( (a, 0) = (2, 0) \)
• \( (-a, 0) = (-2, 0) \)

These vertices indicate where the two branches of the hyperbola open from, lying symmetrically about the y-axis.

Foci of a Hyperbola

The foci of a hyperbola are significant because they help define the shape and orientation of the graph. The two foci are located further away from the center than the vertices and affect the extent of the curve.

For any hyperbola described by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the foci are found using the formula \( (\pm c, 0) \), where \( c = \sqrt{a^2 + b^2} \).

In our hyperbola's equation, this results in \( c = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \), therefore implying foci positions at:

• \( (2\sqrt{5}, 0) \)
• \( (-2\sqrt{5}, 0) \)

These foci are essential in the definition of the hyperbola, showing the points around which the curve seems to stretch.

Asymptotes of a Hyperbola

Asymptotes provide a striking guide as to how steeply the arms of a hyperbola will open. They define invisible boundaries that the hyperbola approaches but never meets, offering insight into the overall shape.

For the given equation \( \frac{x^2}{4} - \frac{y^2}{16} = 1 \), the asymptotes are lines that pass through the center and are not part of the curve itself. Their equations are given by \( y = \pm \frac{b}{a}x \).

Plugging in our values, we have:

• \( y = 2x \)
• \( y = -2x \)

These lines cross at the center \( (0,0) \) and the hyperbola's branches approach these lines as they extend away from the vertices.