Problem 12 - A Graphical Approach to College Algebra

Question

Graph each ellipse by hand. Give the domain and range. Give the foci and identify the center. Do not use a calculator. $$\frac{x^{2}}{16}+\frac{y^{2}}{36}=1$$

Step-by-Step Solution

Answer

Center: (0, 0), Foci: (0, ±2√5), Domain: [-4, 4], Range: [-6, 6].

Step 1: Identify the Standard Form

The given ellipse equation \( \frac{x^2}{16} + \frac{y^2}{36} = 1 \) is already in the standard form of an ellipse centered at the origin \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a^2 = 16 \) and \( b^2 = 36 \).

Step 2: Determine the Center

Since there are no additional terms added or subtracted from \( x \) or \( y \), the center of the ellipse is at the origin, \( (0, 0) \).

Step 3: Calculate the Axes Lengths

The lengths of the axes can be determined by \( a \) and \( b \). Since \( b = 6 \) (from \( b^2 = 36 \)) and \( a = 4 \) (from \( a^2 = 16 \)), the ellipse is vertical with a vertical semi-major axis of length \( 6 \) and a horizontal semi-major axis of length \( 4 \).

Step 4: Graph the Ellipse

To graph the ellipse, plot points at \( (0,\pm6) \) on the vertical axis and \( (\pm4, 0) \) on the horizontal axis. Then sketch the shape of an elliptical curve connecting these points symmetrically around the center.

Step 5: Find the Foci

The foci are found using the formula \( c = \sqrt{b^2 - a^2} \). Here, \( c = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \). The foci are at \( (0, \pm2\sqrt{5}) \).

Step 6: Determine the Domain and Range

The domain of the function, representing how far left and right the ellipse stretches, is \([-4, 4]\). The range, showing how far up and down it stretches, is \([-6, 6]\).

Key Concepts

Domain and Range of an Ellipse

Understanding the domain and range of an ellipse helps us see its extension across the coordinate plane. The domain of an ellipse is all values of \( x \) that lie within the ellipse's horizontal extent. For the equation \( \frac{x^2}{16} + \frac{y^2}{36} = 1 \):

• Since \( a^2 = 16 \), \( a = 4 \). Therefore, the horizontal stretch is from \( -4 \) to \( 4 \).
• The domain is \([-4, 4]\).

The range tells us the vertical stretching of the ellipse. Since \( b^2 = 36 \), \( b = 6 \), it stretches vertically from \(-6 \) to \( 6 \). The range of the ellipse is \([-6, 6]\). This gives a clear picture of the ellipse's span on the plane.

Foci of an Ellipse

The foci are two special points inside the ellipse. The distances from the foci to any point on the ellipse have a constant sum. Finding the foci involves this equation: \[ c = \sqrt{b^2 - a^2} \] Substituting the values we have:

• \( b^2 = 36 \)
• \( a^2 = 16 \)
• Thus, \( c = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5} \)

This calculation shows that the foci are located vertically at \((0, \pm2\sqrt{5})\). These points lie along the major axis, which clarifies why the ellipse is vertically stretched.

Center of an Ellipse

The center of an ellipse is a crucial fixed point equidistant from the extremities of its axes. In our given ellipse equation \( \frac{x^2}{16} + \frac{y^2}{36} = 1 \), the center can be identified by noticing the absence of \( x \) or \( y \) terms being added or subtracted. This lack of shift indicates the center is at the origin, \((0, 0)\).

• This point serves as the balance point of the ellipse, ensuring symmetry vertically and horizontally.
• All calculations for the domain, range, and foci revolve around this central, unmovable point.

Understanding the center's position enhances the grasp of how the ellipse is perfectly symmetrical.