Problem 32 - 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. $$ y^{2}-3 x^{2}=3 $$

Step-by-Step Solution

Answer

Asymptotes: \( y = \pm \sqrt{3}x \); foci at \((0, 2)\) and \((0, -2)\).

Step 1: Rewrite in Standard Form

The given equation is \( y^{2} - 3x^{2} = 3 \). To express this in standard form, divide the entire equation by 3 to get:\[ \frac{y^{2}}{3} - \frac{x^{2}}{1} = 1 \]. This is now in the standard form for a hyperbola \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \) with \( a^{2} = 3 \) and \( b^{2} = 1 \).

Step 2: Identify Asymptotes

For a hyperbola centered at the origin given by \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), the equations for the asymptotes are \( y = \pm \frac{a}{b}x \).Thus, substituting \( a^{2} = 3 \) and \( b^{2} = 1 \), we have \( a = \sqrt{3} \) and \( b = 1 \).Therefore, the asymptotes are \( y = \pm \sqrt{3}x \).

Step 3: Identify and Plot Foci

The distance to each focus from the center of the hyperbola is given by \( c = \sqrt{a^{2} + b^{2}} \).Here, \( c = \sqrt{3 + 1} = 2 \).Thus, the foci are located at \((0, 2)\) and \((0, -2)\), as the hyperbola opens vertically.

Step 4: Sketch the Hyperbola

To sketch the hyperbola:1. Plot the vertices at \((0, \sqrt{3})\) and \((0, -\sqrt{3})\).2. Draw the asymptotes \( y = \sqrt{3}x \) and \( y = -\sqrt{3}x \) as dashed lines.3. Mark the foci at \((0, 2)\) and \((0, -2)\).4. Draw the hyperbola opening upwards and downwards, approaching the asymptotes.

Key Concepts

Standard Form Equation

To truly grasp the nature of hyperbolas, it's important to begin with understanding their standard form equation. The standard form for a hyperbola can take two forms, depending on its orientation. For a hyperbola centered at the origin:

• If it opens vertically, it is represented as: \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \).
• If it opens horizontally, it is \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \).

In the exercise, we transformed the equation \( y^{2} - 3x^{2} = 3 \) into a standard form by dividing through by 3. This gave us \( \frac{y^{2}}{3} - \frac{x^{2}}{1} = 1 \). Here, \( a^{2} \) is 3, and \( b^{2} \) is 1, making \( a = \sqrt{3} \) and \( b = 1 \).
Understanding this transformation helps in identifying key features of the hyperbola such as its axes and orientation.

Asymptotes

Asymptotes play a crucial role in shaping how hyperbolas are graphed. These lines provide a guide that the hyperbola approaches but never touches. For hyperbolas centered at the origin, the equations of the asymptotes depend on the orientation:

• For a vertical hyperbola \( \frac{y^{2}}{a^{2}} - \frac{x^{2}}{b^{2}} = 1 \), the asymptotes are rendered as \( y = \pm \frac{a}{b}x \).
• For a horizontal hyperbola, they are \( y = \pm \frac{b}{a}x \).

In our example, since the hyperbola opens vertically, we calculate the asymptotes using \( a = \sqrt{3}, b = 1 \) yielding \( y = \pm \sqrt{3}x \).
Asymptotes help in drawing the hyperbola and understanding its directionality, as it stretches towards infinity along these lines without ever intersecting them.

Foci of Hyperbola

The foci of a hyperbola add another layer of insight into its geometric properties. These are specific points located along the hyperbola's principal axis, helping to define the curve's "openness". To find the coordinates of the foci:

• Calculate \( c = \sqrt{a^{2} + b^{2}} \).

This formula helps determine how far each focus is from the center.
In our case with \( a^{2} = 3 \) and \( b^{2} = 1 \), we find \( c = \sqrt{3 + 1} = 2 \). Since our hyperbola opens vertically, the foci are located at \((0, 2)\) and \((0, -2)\).
The foci provide insight into the spatial distribution and spread of the arms of the hyperbola, playing a vital role in sketching and analytical geometry.