Problem 9 - Precalculus with Calculus Previews

Question

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ \frac{(x-1)^{2}}{49}+\frac{(y-3)^{2}}{36}=1 $$

Step-by-Step Solution

Answer

Center: \((1, 3)\); Foci: \((1 ± \sqrt{13}, 3)\); Vertices: \((-6, 3), (8, 3)\); Minor Axis Endpoints: \((1, -3), (1, 9)\); Eccentricity: \(\frac{\sqrt{13}}{7}\).

Step 1: Identify the Standard Form

The given equation is \( \frac{(x-1)^2}{49} + \frac{(y-3)^2}{36} = 1 \). This matches the standard form of an ellipse \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), where the center of the ellipse is \((h, k)\), \(a^2\) is the denominator of the fraction associated with \((x-h)^2\), and \(b^2\) is the denominator of the fraction associated with \((y-k)^2\). Compare the given equation to identify these parameters.

Step 2: Determine the Center

For \( \frac{(x-1)^2}{49} + \frac{(y-3)^2}{36} = 1 \), the values \(h = 1\) and \(k = 3\), giving us a center at \((1, 3)\).

Step 3: Identify a and b

Here, \( a^2 = 49 \) and \( b^2 = 36 \). Solving these, we find \( a = 7 \) and \( b = 6 \). Since \(a > b\), the major axis is parallel to the x-axis.

Step 4: Find the Vertices

Vertices are located \(a\) units from the center along the major axis. Thus, the vertices are at \((1 - 7, 3) = (-6, 3)\) and \((1 + 7, 3) = (8, 3)\).

Step 5: Find the Endpoints of the Minor Axis

The minor axis endpoints are \(b\) units from the center along the minor axis. Thus, the endpoints are \((1, 3 - 6) = (1, -3)\) and \((1, 3 + 6) = (1, 9)\).

Step 6: Calculate the Eccentricity

The eccentricity \(e\) of an ellipse is given by \( e = \sqrt{1 - \frac{b^2}{a^2}} \). Substituting \( a^2 = 49 \) and \( b^2 = 36 \): \( e = \sqrt{1 - \frac{36}{49}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7} \).

Step 7: Determine the Foci

The foci are located \(c\) units from the center along the major axis, where \( c = ae = 7\cdot \frac{\sqrt{13}}{7} = \sqrt{13} \). So, the coordinates of the foci are \((1 - \sqrt{13}, 3)\) and \((1 + \sqrt{13}, 3)\).

Key Concepts

Center of Ellipse

The center of an ellipse is a key point that defines its position on the coordinate plane. To find the center, we look at the standard form of the ellipse equation: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here, the center is expressed as \((h, k)\). It's crucial for understanding how the ellipse is oriented.

For the ellipse given by \( \frac{(x-1)^2}{49} + \frac{(y-3)^2}{36} = 1 \), by comparing with the standard form, the center \((h, k)\) comes out to be \((1, 3)\).

This means the entire shape of the ellipse "revolves" around this point. So, when sketching or interpreting graphs, the center acts much like a fixed point or the balancing point of the ellipse.

• The value of \(h\) is the x-coordinate, in this case 1.
• The value of \(k\) is the y-coordinate, in this case 3.

Understanding the center helps place the ellipse correctly in any graphical representation.

Vertices of Ellipse

Vertices are among the most crucial points on the ellipse, located at the ends of the major axis. The major axis is the longest diameter of the ellipse, and it helps to define the overall shape and stretch.

For the given ellipse, the major axis is parallel to the x-axis because \(a^2 = 49\) is larger than \(b^2 = 36\). The formula for the vertices can be derived from the center \((h, k)\) by adding and subtracting \(a\) (which equals 7 in this case) to/from \(h\).

So, the vertices for \( \frac{(x-1)^2}{49} + \frac{(y-3)^2}{36} = 1 \) are:

• \((-6, 3)\) which is \((h-a, k)\)
• \((8, 3)\) which is \((h+a, k)\)

These points represent the farthest distance across the ellipse, highlighting the stretch along the major axis. Visualizing the ellipse with its vertices helps in clarifying the diagram and provides cues for how elongated or circular the ellipse might be.

Eccentricity of Ellipse

Eccentricity \(e\) is a measure of how much an ellipse deviates from being a circle. It’s a value between 0 and 1, where 0 indicates a perfect circle, and values closer to 1 indicate a more elongated shape.

For an ellipse, eccentricity is calculated using the formula \( e = \sqrt{1 - \frac{b^2}{a^2}} \). In this case where \(a^2 = 49\) and \(b^2 = 36\), the eccentricity is calculated as:

\[ e = \sqrt{1 - \frac{36}{49}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7} \]

This value, \(\frac{\sqrt{13}}{7}\), tells us that the ellipse is slightly elongated along the x-axis. The closer the eccentricity is to zero, the rounder the ellipse. Understanding eccentricity is important because it helps in predicting the level of deviation of the ellipse from being circular. This makes it a significant parameter when describing physical and astronomical ellipses.

Foci of Ellipse

Foci are two special focus points inside the ellipse which hold a critical property: for any point on the ellipse, the sum of the distances to the two foci is constant. The location of the foci helps in defining the shape and size of the ellipse.

For the given example, using the eccentricity \(e = \frac{\sqrt{13}}{7}\), we find \(c\) (the distance from the center to each focus) by the multiplication \(c = ae\). Here, \(a = 7\), thus \(c = 7 \times \frac{\sqrt{13}}{7} = \sqrt{13}\).

The coordinates of the foci are determined by moving \(c\) units from the center along the major axis, which in this equation is along the x-axis. Thus, the foci are:

• \((1 - \sqrt{13}, 3)\)
• \((1 + \sqrt{13}, 3)\)

These points ensure that the unique property of the ellipse (constant sum of distances) is maintained. Understanding the role of foci is often essential in problems involving light, sound, and other physics concepts where ellipses play a role.