Problem 50 - A Graphical Approach to Precalculus with Limits

Question

CHECKING ANALYTIC SKILLS Graph each hyberbola by hand. Give the domain and range. Give the center in Exercises \(55-61 .\) Do not use a calculator. $$144 x^{2}-49 y^{2}=7056$$

Step-by-Step Solution

Answer

Center: (0, 0); Domain: (-∞, -7) ∪ (7, ∞); Range: (-∞, ∞).

Step 1: Write the Equation in Standard Form

The given equation is \(144x^2 - 49y^2 = 7056\). Divide every term by 7056 to rewrite the equation in standard form:\[\frac{144x^2}{7056} - \frac{49y^2}{7056} = 1\]Simplify each term to get:\[\frac{x^2}{49} - \frac{y^2}{144} = 1\]This represents a hyperbola centered at the origin, opening along the x-axis.

Step 2: Identify the Center

Since the equation \(\frac{x^2}{49} - \frac{y^2}{144} = 1\) is in standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\) with \((h, k) = (0, 0)\), the center of the hyperbola is at the point \((0, 0)\).

Step 3: Determine the Transverse and Conjugate Axes

For \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) where \(a^2 = 49\) (thus \(a = 7\)) and \(b^2 = 144\) (thus \(b = 12\)), the transverse axis is along the x-axis and its length is \(2a = 14\). The conjugate axis is along the y-axis with length \(2b = 24\).

Step 4: Sketch the Hyperbola

From the center \((0, 0)\), plot points at \(x = \pm a (\pm7)\) on the transverse axis and \(y = \pm b (\pm12)\) on the conjugate axis. These points are vertices and co-vertices respectively. Draw the box around these points and sketch the asymptotes that pass diagonally through the box. Finally, draw the hyperbola opening horizontally along the transverse axis.

Step 5: Determine the Domain and Range

Since the hyperbola opens horizontally, the domain covers all real numbers that satisfy \(x^2 > 49\), or \(x > 7\) and \(x < -7\), meaning the domain is \((-fty, -7) \cup (7, \infty)\). The range of a hyperbola that is oriented horizontally includes all real numbers, so the range is \((-fty, fty)\).

Key Concepts

Standard Form of Hyperbola

Understanding the standard form of a hyperbola is a key step in graphing and analyzing these fascinating curves. A hyperbola can be expressed in the standard form as either:

• For horizontal opening: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• For vertical opening: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Here, \((h, k)\) represents the center of the hyperbola. The values \(a^2\) and \(b^2\) determine the distances from the center to the vertices and co-vertices, respectively.
If a hyperbola is given in a general form like \(Ax^2 - By^2 = C\), you first need to divide each term by \(C\) to transform it into the standard form. Simplifying the given equation \(144x^2 - 49y^2 = 7056\) into standard form, you get \(\frac{x^2}{49} - \frac{y^2}{144} = 1\), indicating that our hyperbola opens horizontally since \((x-h)^2\) comes first.

Transverse and Conjugate Axes

Once the hyperbola equation is in standard form, you can easily identify the transverse and conjugate axes. These axes affect how the hyperbola is oriented and its overall shape.

• Transverse Axis: Lies along the direction where \((x-h)^2\) or \((y-k)^2\) has the positive coefficient. For our example \(\frac{x^2}{49} - \frac{y^2}{144} = 1\), it stretches horizontally along the x-axis, making the transverse axis horizontal.
• Conjugate Axis: Perpendicular to the transverse axis and helps define the "width" of the hyperbola. This axis is vertical in our example.

The length of these axes are given by:

• Transverse axis length: \(2a = 14\) (since \(a = 7\))
• Conjugate axis length: \(2b = 24\) (since \(b = 12\))

Using these axes, you can mark the vertices on the transverse axis at \((\pm7, 0)\) and the co-vertices on the conjugate axis at \((0, \pm12)\). These points assist in graphing the hyperbola.

Domain and Range of Hyperbolas

The domain and range of a hyperbola can be a bit tricky since these figures stretch out infinitely, but understanding their basic definitions makes this simpler.

• Domain: For a hyperbola opening horizontally, like \(\frac{x^2}{49} - \frac{y^2}{144} = 1\), the possible values of \(x\) are where the hyperbola actually exists, which occurs when \(x^2 > 49\). This implies the domain is all real numbers except for the interval between \(-7\) and \(7\). Therefore, the domain is \((-\infty, -7) \cup (7, \infty)\).
• Range: Similarly, for horizontal hyperbolas, the range includes all real numbers because the hyperbola can rise or fall indefinitely: \((-\infty, \infty)\).

Remember, the hyperbola extends in regions beyond the vertices on the horizontal plane, and thus covers the entire set of all real numbers vertically within its range.