Problem 12 - Precalculus: Mathematical for Calculus

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$x^{2}-y^{2}+4=0$$

Step-by-Step Solution

Answer

Vertices: (2, 0), (-2, 0); Foci: (2√2, 0), (-2√2, 0); Asymptotes: y = ±x.

Step 1: Rewrite the Equation in Standard Hyperbola Form

The standard form of a hyperbola with a horizontal transverse axis is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \). Start by isolating the quadratic parts of the equation.Given: \[ x^2 - y^2 + 4 = 0 \]Rewriting, we have: \[ x^2 - y^2 = -4 \]Divide all terms by \(-4\) to obtain: \[ \frac{x^2}{4} - \frac{y^2}{4} = 1 \]This equation shows a hyperbola with its center at the origin \((h, k) = (0, 0)\).

Step 2: Identify Vertices, Foci, and Asymptotes

In the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the values of \(a\) and \(b\) relate to the hyperbola's dimensions.From \( \frac{x^2}{4} \), we obtain \( a^2 = 4 \) which gives \( a = 2 \), and from \(-\frac{y^2}{4} \), we have \( b^2 = 4 \) giving \( b = 2 \).The vertices, located at \((h\pm a, k) = (\pm 2, 0)\), are at the points \((2, 0)\) and \((-2, 0)\).The foci are given by \(c\) where \(c^2 = a^2 + b^2 = 4 + 4 = 8 \), so \(c = \sqrt{8} = 2\sqrt{2}\). The foci are at \((\pm 2\sqrt{2}, 0)\).The asymptotes of the hyperbola are given by \( y = \pm\frac{b}{a}x \), hence \( y = \pm x \).

Step 3: Sketch the Graph

Plot the center, vertices, foci, and asymptotes on the graph.1. Center at origin \((0, 0)\).2. Plot vertices at \((2, 0)\) and \((-2, 0)\).3. Plot the foci at \((2\sqrt{2}, 0)\) and \((-2\sqrt{2}, 0)\).4. Draw the asymptotes \( y = x \) and \( y = -x \).5. Sketch the branches of the hyperbola opening left and right between the asymptotes.

Key Concepts

Vertices of a Hyperbola

The vertices of a hyperbola are crucial as they help define the shape and width of each branch of the hyperbola. For a hyperbola centered at the origin with a horizontal transverse axis, the vertices lie along the x-axis. The general form for such a hyperbola is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] where \((h, k)\) is the center of the hyperbola, and \(a\) is the distance from the center to each vertex along the transverse axis. In our specific case, the hyperbola given by the equation: \[ \frac{x^2}{4} - \frac{y^2}{4} = 1 \]has a center at the origin \((0, 0)\), and \(a^2 = 4\), so \(a = 2\). This indicates the vertices are at \((h\pm a, k)\), leading to the coordinates \((2, 0)\) and \((-2, 0)\).To better visualize:

• The center is at the origin, \((0, 0)\).
• Vertices are symmetrically placed on the transverse axis.

These vertices mark the broadest part of the hyperbola branches, lying equidistant from the center.

Foci of a Hyperbola

The foci of a hyperbola are two fixed points located along the transverse axis that help define its curvature and eccentricity. For any hyperbola, each point on the curve has the sum of its distances to the foci being constant. This property uniquely defines hyperbolas.In our example, the hyperbola is centered at \((0, 0)\), with a transverse axis along the x-axis:\[ \frac{x^2}{4} - \frac{y^2}{4} = 1 \]The formula for finding the foci for a standard horizontal hyperbola is given by \(c^2 = a^2 + b^2\), where \(a^2 = 4\) and \(b^2 = 4\). Therefore, solving gives \(c^2 = 8\), which means \(c = \sqrt{8} = 2\sqrt{2}\).Thus, the coordinates of the foci are:

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

These points lie further out than the vertices, and are used when plotting the hyperbola to ensure accuracy.

Asymptotes of a Hyperbola

Asymptotes for hyperbolas are important as they provide guidelines that the branches of the hyperbola approach but never actually touch. These are diagonal lines that extend infinitely and are critical in sketching a proper hyperbola.For a hyperbola of the form:\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]centered at \((h, k)\), the equations of the asymptotes are given by \(y - k = \pm \frac{b}{a}(x - h)\). For the equation \(\frac{x^2}{4} - \frac{y^2}{4} = 1\), with the center at \((0, 0)\), this results in:\[ y = \pm \frac{b}{a}x \]Given \(a = 2\) and \(b = 2\), the asymptotes simplify to the lines:

• \(y = x\)
• \(y = -x\)

These lines pass through the origin and provide a visual framework within which to sketch the hyperbola's branches. Recognizing the role asymptotes play is essential in understanding the behavior and shape of hyperbolas.