Problem 10 - Contemporary Precalculus

Question

Identify the conic section whose equation is given and find its graph. List its vertices, foci, and asymptotes. $$4 x^{2}-y^{2}=16$$

Step-by-Step Solution

Answer

Question: Determine the type of conic section represented by the equation \(16 - 16y^2 = x^2\). Find its vertices, foci, and asymptotes if applicable. Answer: The given equation represents a hyperbola. The vertices are at points (4, 0) and (-4, 0). The foci are at points (4√3, 0) and (-4√3, 0). The asymptotes are given by the lines y = ±√2x.

Step 1: Identify the conic section

To identify whether the given equation represents an ellipse or a hyperbola, we need to rewrite the equation in the standard form. Divide both sides of the equation by 16 to obtain the standard form: $$\frac{x^2}{4^2} - \frac{y^2}{(4\sqrt{2})^2} = 1$$ The equation has the form $$\frac{x^2}{a^2} - \frac{y^2}{b^2}=1$$ which is the standard form of a hyperbola where $$a=4$$ and $$b=4\sqrt{2}$$. Consequently, the given equation represents a hyperbola.

Step 2: Find the vertices

The vertices of the hyperbola are located at points $$(\pm a, 0)$$, where $$a=4$$. Thus, the vertices are at points $$(4, 0)$$ and $$(-4, 0)$$.

Step 3: Find the foci

The foci of the hyperbola are located at points $$(\pm c, 0)$$, where $$c=\sqrt{a^2+b^2}$$. With $$a=4$$ and $$b=4\sqrt{2}$$, we have: $$c = \sqrt{4^2 + (4\sqrt{2})^2} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}$$ Thus, the foci are at points $$(4\sqrt{3}, 0)$$ and $$(-4\sqrt{3}, 0)$$.

Step 4: Find the asymptotes

The asymptotes of the hyperbola are given by the lines $$\frac{y}{b} = \pm \frac{x}{a}$$. In this case, we have $$a=4$$ and $$b=4\sqrt{2}$$, so the asymptotes are the lines: $$\frac{y}{4\sqrt{2}} = \pm \frac{x}{4}$$ which can be rewritten as $$y = \pm \sqrt{2}x$$. To summarize: - The given equation represents a hyperbola. - Vertices: $$(4, 0)$$ and $$(-4, 0)$$ - Foci: $$(4\sqrt{3}, 0)$$ and $$(-4\sqrt{3}, 0)$$ - Asymptotes: $$y = \pm \sqrt{2}x$$

Key Concepts

Conic Sections

Conic sections, a foundational concept in geometry and algebra, are shapes created by slicing a right circular cone with a plane at different angles. Depending on the angle of the cut, we can get one of the four types of conics: circles, ellipses, parabolas, and hyperbolas. Each conic section has a distinct set of equations and properties that define its shape and characteristics.

For example, the standard form of a hyperbola's equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) if the transverse axis is horizontal, or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) if the transverse axis is vertical, where \( a \) and \( b \) are the hyperbola's real semiaxes. Understanding the equation's structure helps us graph the shape and identify its key features, such as its vertices, foci, and asymptotes.

Vertices of a Hyperbola

The vertices of a hyperbola are points where the hyperbola intersects its transverse axis. This axis is the segment that runs through the center of the hyperbola, connecting the vertices and lying between them. For the given hyperbola with the equation, \( 4x^2 - y^2 = 16 \), we calculate the vertices by setting \( y \) to zero and solving for \( x \) in terms of \( a \).

The vertices are identified as \( (\pm a, 0) \) where \( a = 4 \) in this case, hence we have the vertices at \( (4, 0) \) and \( (-4, 0) \) for the hyperbola. The position of these vertices gives us a visual cue to the hyperbola's orientation and width.

Foci of a Hyperbola

The foci of a hyperbola are a pair of points located along the transverse axis, both equidistant from the center of the hyperbola. They are an intrinsic part of a hyperbola's definition and its geometric shape. For a hyperbola centered at the origin, the distance of each focus from the center is denoted by \( c \), which is calculated by the relationship \( c^2 = a^2 + b^2 \).

In our hyperbola \( 4x^2 - y^2 = 16 \), we find \( c \) by substituting the values of \( a \) and \( b \) yielding \( c = \sqrt{4^2 + (4\sqrt{2})^2} = 4\sqrt{3} \). Thus, the foci are at \( (4\sqrt{3}, 0) \) and \( (-4\sqrt{3}, 0) \). The foci play a crucial role when constructing the graph of the hyperbola as their location helps determine the shape's spread or eccentricity.

Asymptotes

Asymptotes are straight lines that a hyperbola's branches approach indefinitely but never actually reach. The asymptotes provide a 'skeleton' that outlines the general form of the hyperbola. For the equation \( 4x^2 - y^2 = 16 \), you can derive the equations of the asymptotes by rewriting the original equation to concur with the formula \( \frac{y}{b} = \pm \frac{x}{a} \) where \( a = 4 \) and \( b = 4\sqrt{2} \). This generates the lines \( y = \pm \sqrt{2}x \).

These asymptotes do not intersect the hyperbola itself but they cross at the center of the hyperbola, guiding us in sketching the open, curving arms of the graph. Asymptotes are essential in accurately graphing a hyperbola as they delineate the bounds within which the hyperbola exists.