Problem 2 - Precalculus: Mathematical for Calculus

Question

Find the center, foci, and vertices of the ellipse, and determine the lengths of the major and minor axes. Then sketch the graph. $$\frac{(x-3)^{2}}{16}+(y+3)^{2}=1$$

Step-by-Step Solution

Answer

Center: (3, -3). Foci: \((3 \pm \sqrt{15}, -3)\). Vertices: \((-1, -3), (7, -3)\). Major axis length: 8. Minor axis length: 2.

Step 1: Identify the Standard Form of an Ellipse

The given equation is \( \frac{(x-3)^2}{16} + (y + 3)^2 = 1 \). This represents an ellipse equation where the sum of two squared terms is equal to 1. Compare this with the standard form of an ellipse \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).

Step 2: Find the Center of the Ellipse

From the equation, \( (x-3)^2 \) and \( (y+3)^2 \) tell us the center of the ellipse is at \( (h, k) = (3, -3) \).

Step 3: Determine the Values of \(a\) and \(b\)

From the denominator of the \(x\) term, \(a^2 = 16\), which gives \(a = 4\). The \(y\) term has an implicit \(b^2 = 1\), so \(b = 1\).

Step 4: Classify the Axes

Since \(a^2 > b^2\) (\(16 > 1\)), the major axis is along the x-direction, and the minor axis is along the y-direction.

Step 5: Find the Vertices

The vertices along the major axis (horizontal) are at \((h \pm a, k)\). So, the vertices are \((3 \pm 4, -3) = (-1, -3)\) and \((7, -3)\).

Step 6: Calculate the Lengths of the Axes

The length of the major axis is \(2a = 8\), and the length of the minor axis is \(2b = 2\).

Step 7: Determine the Foci

The foci are found using \(c^2 = a^2 - b^2\). Here, \(c^2 = 16 - 1 = 15\), so \(c = \sqrt{15}\). The foci are at \((h \pm c, k) = (3 \pm \sqrt{15}, -3)\).

Step 8: Sketch the Graph

Plot the center at \((3, -3)\). Draw a horizontal major axis of length 8 and a vertical minor axis of length 2 intersecting at the center. Mark the vertices at \((-1, -3)\) and \((7, -3)\), and the foci at \((3 - \sqrt{15}, -3)\) and \((3 + \sqrt{15}, -3)\).

Key Concepts

Center of Ellipse

The center of an ellipse is a crucial point that serves as its midpoint, helping us understand its overall structure. In the equation of an ellipse that resembles the form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), the center is located at \((h, k)\). The given equation \( \frac{(x-3)^2}{16} + (y + 3)^2 = 1 \) shows us that the center is at \((3, -3)\) because the expressions \((x-3)^2\) and \((y+3)^2\) suggest adjustments or shifts from the origin. To break it down further:

• \(x-3\) indicates a horizontal shift 3 units to the right.
• \(y+3\) indicates a vertical shift 3 units downward.

So, the point \((3, -3)\) is where the ellipse is centered in the Cartesian plane.

Vertices of Ellipse

Vertices are points that define the extremities of the ellipse along its major axis. For the ellipse equation \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), the vertices are found by moving \(a\) units from the center along the major axis. In the provided equation, we've determined \(a = 4\) and \(b = 1\). Since \(a > b\), the major axis is horizontal.The vertices can be calculated as:

• \((h + a, k) = (3 + 4, -3) = (7, -3)\)
• \((h - a, k) = (3 - 4, -3) = (-1, -3)\)

These points, \((7, -3)\) and \((-1, -3)\), represent the farthest horizontal points of the ellipse.

Major and Minor Axes

The major and minor axes are the longest and shortest diameters of an ellipse, respectively. These axes help us visualize the orientation and shape of the ellipse.- **Major Axis**: It's aligned with the largest value of \(a\). In our equation, \(a = 4\) and \(b = 1\), so the major axis is horizontal, along the \(x\)-axis. Its full length is \(2a = 8\) units.- **Minor Axis**: It runs perpendicular to the major axis. Here, the minor axis is vertical with a length of \(2b = 2\) units.The major and minor axes intersect at the center, \((3, -3)\). This configuration makes the ellipse wider than it is tall, indicating a horizontal stretch.

Foci of Ellipse

Foci are special points on the major axis, inside the ellipse, crucial for its geometric properties. An ellipse is defined such that the sum of the distances from any point on the ellipse to the two foci is a constant.For our ellipse:- Use the formula for the distance to the foci: \(c^2 = a^2 - b^2\).- Here, \(a^2 = 16\) and \(b^2 = 1\), leading to \(c^2 = 15\).- Hence, \(c = \sqrt{15}\).The foci are positioned along the major axis, away from the center:

• \((3 + \sqrt{15}, -3)\)
• \((3 - \sqrt{15}, -3)\)

These points are important as they maintain the constant sum of distances property that characterizes ellipses.