Problem 31 - Thomas Calculus

Question

Exercises \(27-34\) give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. $$ 8 x^{2}-2 y^{2}=16 $$

Step-by-Step Solution

Answer

The hyperbola in standard form is \(\frac{x^2}{2} - \frac{y^2}{8} = 1\), with asymptotes \(y = \pm 2x\), and foci at \((\pm \sqrt{10}, 0)\).

Step 1: Write the given equation in standard form

Start with the given equation of the hyperbola: \(8x^2 - 2y^2 = 16\). Divide both sides by 16 to begin converting it to standard form. The result will be \(\frac{x^2}{2} - \frac{y^2}{8} = 1\).

Step 2: Identify the standard form of the hyperbola

The equation \(\frac{x^2}{2} - \frac{y^2}{8} = 1\) represents a hyperbola that has the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Here, \(a^2 = 2\) and \(b^2 = 8\).

Step 3: Calculate asymptotes of the hyperbola

The asymptotes for a hyperbola centered at the origin of the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are given by the equations \(y = \pm \frac{b}{a}x\). Substituting \(a^2 = 2\) and \(b^2 = 8\) to find \(a = \sqrt{2}\) and \(b = 2\sqrt{2}\), the asymptotes are \(y = \pm 2x\).

Step 4: Locate the foci of the hyperbola

The distance from the center to each focus is given by \(c = \sqrt{a^2 + b^2}\). Plug in \(a^2 = 2\) and \(b^2 = 8\), so \(c = \sqrt{10}\). The foci are located at \((\pm \sqrt{10}, 0)\).

Step 5: Sketch the hyperbola with asymptotes and foci

Draw the coordinate axes. Plot the center at the origin, (0,0). Mark the vertices at \((\pm \sqrt{2}, 0)\), draw the asymptotes with slopes \(\pm 2\), and finally place the foci at \((\pm \sqrt{10}, 0)\). Sketch the hyperbola opening left and right between these vertices, approaching the asymptotes.

Key Concepts

Asymptotes

In a hyperbola, asymptotes are straight lines that the curve approaches but never touches. They guide the shape of the hyperbola as it extends to infinity. For a hyperbola centered at the origin and in the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm \frac{b}{a}x\). These lines provide a boundary that the hyperbola cannot cross.

• First, identify \(a^2\) and \(b^2\) from the hyperbola's equation.
• Then calculate \(a = \sqrt{a^2}\) and \(b = \sqrt{b^2}\).
• The slopes of the asymptotes are then \(\pm \frac{b}{a}\).

In the given example, \(a = \sqrt{2}\) and \(b = 2\sqrt{2}\), giving asymptotes as \(y = \pm 2x\). Asymptotes help in sketching and understanding the hyperbola's orientation and spread.

Foci

The foci of a hyperbola are two specific points that play a crucial role in its definition. These points are located along the transverse axis, which is the axis that passes through the vertices of the hyperbola.
The distance from the center of the hyperbola to each focus is denoted by \(c\), calculated by the formula \(c = \sqrt{a^2 + b^2}\). This distance is always greater than the distance to the vertices, ensuring the foci are outside the 'box' formed by the asymptotes.
For our hyperbola example, \(c = \sqrt{10}\), meaning the foci are found at \((\pm \sqrt{10}, 0)\). These points indicate the direction in which the hyperbola opens and provide important reference points for both graphing and understanding its geometric properties.

Sketching Hyperbolas

Sketching a hyperbola involves several steps that make use of its geometrical properties and characteristics.
Start by plotting the center of the hyperbola, often at the origin in standard equations. Then, mark the vertices, which are located \(a\) units from the center along the transverse axis (for our example, at \((\pm \sqrt{2}, 0)\)).
Next, draw the asymptotes through the center with the slope calculated earlier. In this situation, these lines have slopes of \(\pm 2\) and create a guiding framework for sketching the curve.
With the foci located and vertices plotted, you can now draw the hyperbola. The branches of the curve will start near the vertices and approach, but never touch, the asymptotes.

• The curve opens outwards from the vertices towards the foci.
• The asymptotes direct the overall shape and the curve's spread.

Consider symmetry and shape as you sketch: the hyperbola will have a symmetric path about both the transverse and conjugate axes.