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

Standard form: \(\frac{y^2}{3} - \frac{x^2}{1} = 1\). Asymptotes: \(y = \pm \sqrt{3}x\).

Step 1: Write the Equation in Standard Form

The given equation is \( y^2 - 3x^2 = 3 \). To write this in standard form, divide every term by \( 3 \) to get:\[ \frac{y^2}{3} - \frac{x^2}{1} = 1 \]This is now in the standard form for a hyperbola with the transverse axis along the y-axis.

Step 2: Identify the Hyperbola's Semiaxes

From the equation \( \frac{y^2}{3} - \frac{x^2}{1} = 1 \), we identify:- \( a^2 = 3 \), so \( a = \sqrt{3} \). This corresponds to the distance from the center to each vertex along the y-axis.- \( b^2 = 1 \), so \( b = 1 \). This corresponds to the distance from the center along the x-axis.

Step 3: Determine the Asymptotes

For hyperbolas of the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the equations of the asymptotes are:\[ y = \pm \frac{a}{b}x \]Substituting \( a = \sqrt{3} \) and \( b = 1 \), we get the asymptotes:\[ y = \pm \sqrt{3}x \]

Step 4: Find the Foci

The foci are located at a distance \( c \) from the center, where \( c^2 = a^2 + b^2 \).Calculating \( c \):\[ c^2 = 3 + 1 = 4 \]\( c = 2 \), so the foci are at \((0, \pm 2)\).

Step 5: Sketch the Hyperbola

On a coordinate plane:1. Draw the central point at the origin \((0,0)\).2. Plot the vertices along the y-axis at \((0, \pm \sqrt{3})\).3. Draw the asymptotes through the origin with slopes \( \pm \sqrt{3} \).4. Mark the foci at \((0, \pm 2)\).5. Sketch the hyperbola opening upwards and downwards, approaching the asymptotes.

Key Concepts

Understanding the Standard Form of Hyperbolas

When we talk about hyperbolas, the standard form of their equation plays a crucial role in identifying the properties and shape of the curve. For a hyperbola centered at the origin, the equation can appear in one of two standard forms:

• \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) when the transverse axis is horizontal.
• \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) when the transverse axis is vertical.

Determining the standard form from a given equation requires rearranging it such that it resembles one of the above structures. This involves dividing through by a common factor to isolate the relationship between the squared terms and setting the equation equal to 1.
In this particular exercise, the equation \( y^2 - 3x^2 = 3 \), after dividing every term by 3, becomes \( \frac{y^2}{3} - \frac{x^2}{1} = 1 \). This identifies the structure where the transverse axis is along the y-axis, indicating that the hyperbola opens upwards and downwards.

Asymptotes of Hyperbolas

Asymptotes are lines that a hyperbola approaches but never intersects. In a hyperbola defined by the standard form, the equations for the asymptotes can be derived based on the orientation of the transverse axis.
For the vertical transverse axis ( \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)), the asymptotes have the equations:

• \( y = \pm \frac{a}{b}x \)

These lines serve as a boundary, showing the path along which the hyperbola will extend outward from the vertices in an open-ended curve.
In the specific exercise, substituting \( a = \sqrt{3} \) and \( b = 1 \), gives us the asymptotes: \( y = \pm \sqrt{3}x \). The slopes of these lines reflect how steeply the hyperbola extends from its vertices, framing the region within which the hyperbola lies.

Locating the Foci of a Hyperbola

The foci of a hyperbola are specific points located along the transverse axis, from which the distances maintaining constant differences define the shape of the curve. To find these points, we employ the relationship:

• \( c^2 = a^2 + b^2 \)

Here, \( c \) is the distance from the center to each focus of the hyperbola.
Substituting the values from our standard form into this equation, we calculate \( c \): \[ c^2 = 3 + 1 = 4 \] Therefore, \( c = 2 \). This tells us that the foci are located at \( (0, \pm 2) \), lying on the y-axis since the transverse axis is vertical. These foci are pivotal in determining the elongated shape of the hyperbola, as they are part of the defining property that differentiates hyperbolas from other conics like ellipses.