Problem 30 - 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. $$ y^{2}-x^{2}=4 $$

Step-by-Step Solution

Answer

The hyperbola's asymptotes are \(y=\pm x\), and the sketch includes these along with foci at \((0,\pm 2.83)\).

Step 1: Identify the hyperbola equation

Given the equation is \(y^2 - x^2 = 4\). Identify this as the equation of a hyperbola where the \(y\) term is positive, suggesting a vertical orientation.

Step 2: Rewrite in standard form

Divide each term by 4 to rewrite the equation in standard form: \[ \frac{y^2}{4} - \frac{x^2}{4} = 1 \]This represents a hyperbola centered at (0,0) with vertical transverse axis.

Step 3: Find the asymptotes

For the hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the equations for the asymptotes are \(y = \pm \frac{a}{b}x\). Here, \(a^2 = 4\) and \(b^2 = 4\), so the asymptotes are \(y = \pm x\).

Step 4: Determine vertices and foci

The vertices, located at \(\pm a\) from the center along the transverse axis, are at points (0, 2) and (0, -2). The foci are at a distance \(c = \sqrt{a^2 + b^2} = \sqrt{8}\) from the center, roughly (0, \(\pm2.83\)).

Step 5: Sketch the hyperbola

Draw the coordinate axes. Plot the center at (0,0). Draw the vertices at (0,2) and (0,-2). Plot the asymptotes \(y=\pm x\) using dashed lines. Sketch the hyperbola opening upwards and downwards, approaching but not touching the asymptotes. Mark foci at \( (0, \pm 2.83) \).

Key Concepts

Asymptotes

In the context of hyperbolas, asymptotes play a crucial role. They are lines that the hyperbola approaches but never actually meets. These lines help define the hyperbola's shape and are essential in sketching its graph.
For the hyperbola given by the equation, \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the asymptotes can be found using the formula:

• \( y = \pm \frac{a}{b}x \)

This equation indicates that the slope of the asymptotes is determined by the ratio \( \frac{a}{b} \). When you place the hyperbola in standard form, you can immediately determine these lines.
In our example, \( a^2 = 4 \) and \( b^2 = 4 \), so both \( a \) and \( b \) equal 2. Therefore, \( \frac{a}{b} = 1 \), leading to the equations \( y = \pm x \) for the asymptotes. These linear equations guide the hyperbola's path as it opens out along the vertical axis.

Standard Form

Converting a hyperbola's equation into standard form is essential for easy interpretation and graphing. The standard form for a hyperbola aligned with the coordinate axes is either

• \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)
• or \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).

The choice between these forms depends on the hyperbola's orientation.
In the equation given, \( y^2 - x^2 = 4 \), you can identify it as a vertical hyperbola. To convert it to the standard form, divide the entire equation by 4:\[ \frac{y^2}{4} - \frac{x^2}{4} = 1 \]
This conversion reveals important information like the values of \( a^2 \) and \( b^2 \), which are both 4 in this case. This means the hyperbola is centered around the origin (0,0) and opens vertically.

Foci

Foci are fundamental points for any hyperbola, defining its precise shape and the direction it opens. For hyperbolas, the foci are located along the transverse axis, a key characteristic that differentiates them from ellipses.
For the hyperbola in standard form, \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the foci are calculated using the formula:

• \( c = \sqrt{a^2 + b^2} \)

The value \( c \) represents the distance from the center to each focus. In our example, \( a^2 = 4 \) and \( b^2 = 4 \), so \( c = \sqrt{8} \approx 2.83 \). Thus, the foci are located at approximately (0, \( \pm 2.83 \)).
This means for this vertically oriented hyperbola, the foci lie along the y-axis, further away than the vertices. Including the foci in your sketch aids in understanding the hyperbola's opening direction and overall structure.