Problem 23 - Precalculus: Mathematics for Calculus

Question

An equation of a hyperbola is given. (a) Find the vertices, foci, and asymptotes of the hyperbola. (b) Determine the length of the transverse axis. (c) Sketch a graph of the hyperbola. $$x^{2}-y^{2}+4=0$$

Step-by-Step Solution

Answer

Vertices: (2,0), (-2,0); Foci: (2\sqrt{2},0), (-2\sqrt{2},0); Asymptotes: y=x, y=-x; Transverse axis length: 4.

Step 1: Rewrite the equation in standard form

The given equation is \( x^{2} - y^{2} + 4 = 0 \). To express it in the standard hyperbola form, rearrange the terms: \( x^{2} - y^{2} = -4 \). By dividing the entire equation by -4, we get: \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \), which is now in the standard form of a hyperbola centered at the origin (\(0,0\)) with \( a^2 = 4 \) and \( b^2 = 4 \).

Step 2: Find the vertices

The vertices of the hyperbola can be found using the formula \( (\pm a, 0) \) where \( a = \sqrt{4} = 2 \). Therefore, the vertices are \( (2, 0) \) and \( (-2, 0) \).

Step 3: Find the foci

The distance to the foci from the center is found using \( c = \sqrt{a^2 + b^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \). The foci are located at \( (\pm 2\sqrt{2}, 0) \).

Step 4: Determine the asymptotes

The asymptotes for this hyperbola are given by the equations \( y = \pm \frac{b}{a}x \). With \( a = 2 \) and \( b = 2 \), the equations are \( y = \pm x \).

Step 5: Length of the transverse axis

The length of the transverse axis is \( 2a \), where \( a = 2 \). Thus, the length is \( 4 \).

Step 6: Sketch the graph

To sketch the hyperbola, plot the center at the origin \((0,0)\), the vertices at \((2,0)\) and \((-2,0)\), the foci at \((2\sqrt{2},0)\) and \((-2\sqrt{2},0)\). Draw the asymptotes as lines \( y = x \) and \( y = -x \). The hyperbola opens left and right along the x-axis.

Key Concepts

Vertices

In the context of hyperbolas, vertices are key points where each branch of the hyperbola makes its closest approach to the center. For a hyperbola centered at the origin, the vertices can be found using the equation's standard form.

Let's consider the equation of our hyperbola: \( \frac{x^2}{4} - \frac{y^2}{4} = 1 \). This equation indicates a hyperbola that opens horizontally, as the \(x^2\) term is positive. The standard formula for a hyperbola that opens horizontally is \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), centered at \((h, k)\). Here, \(h = 0\) and \(k = 0\), so the center is \((0, 0)\).

The distance from the center to each vertex is \(a\), where \(a^2 = 4\). Taking the square root, we find \(a = 2\). Thus, the vertices are at \((\pm 2, 0)\). This tells us the branches of the hyperbola are closest to the center at these points, spaced symmetrically along the x-axis.

Foci

The foci of a hyperbola are integral to its geometric definition. They are two fixed points situated inside each branch, and the distance between any point on the hyperbola and each focus remains constant. This unique property defines the hyperbola's shape.

To determine the foci, we use the formula \(c = \sqrt{a^2 + b^2}\), where \(a\) and \(b\) are derived from the hyperbola's equation. For our equation \(\frac{x^2}{4} - \frac{y^2}{4} = 1\), we already established \(a^2 = 4\) and \(b^2 = 4\). Plugging these into the formula gives us:

• \(c = \sqrt{4 + 4} = \sqrt{8} \)
• Simplifying further, \(c = 2\sqrt{2} \)

The foci are placed at \((\pm 2\sqrt{2}, 0)\), equidistant from the center along the x-axis. These points do not lie on the hyperbola itself but inside its arms, guiding its curvature.

Asymptotes

Asymptotes are crucial to understanding the infinite nature of hyperbolas. These lines, though not touching the hyperbola, show how the arms of the hyperbola behave as they extend further away from the center. For a hyperbola, the asymptotes pass through the center, providing essential visual guides.

The asymptotes of a hyperbola are given by the equation \(y = \pm \frac{b}{a}x\). In our hyperbola \(\frac{x^2}{4} - \frac{y^2}{4} = 1\), both \(a = 2\) and \(b = 2\). This means the relationship simplifies to:

• Asymptote equations: \(y = \pm \frac{2}{2}x = \pm x\)

These lines, \(y = x\) and \(y = -x\), create a cross intersecting at the origin. The arms of the hyperbola approach these lines as they go towards infinity, reflecting the asymptotic behavior inherent in hyperbolas.