Problem 10 - Precalculus with Calculus Previews

Question

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ \frac{(x+1)^{2}}{25}+\frac{(y-2)^{2}}{36}=1 $$

Step-by-Step Solution

Answer

The center is (-1, 2), vertices at (-1, 8) and (-1, -4), co-vertices at (4, 2) and (-6, 2), foci at (-1, 2 ± √11), and eccentricity is √11/6.

Step 1: Identify the Standard Form of the Ellipse

The equation \( \frac{(x+1)^{2}}{25} + \frac{(y-2)^{2}}{36} = 1 \) is in the form \( \frac{(x-h)^{2}}{b^{2}} + \frac{(y-k)^{2}}{a^{2}} = 1 \), where \(a > b\). Here, \(a^{2} = 36\) and \(b^{2} = 25\). Thus, we have \(a = 6\) and \(b = 5\). The center of the ellipse is \((h, k) = (-1, 2)\).

Step 2: Determine the Vertices

The vertices are located \(a\) units from the center along the major axis. Since the major axis is vertical (because \(a > b\)), the vertices are at \((-1, 2 + 6)\) and \((-1, 2 - 6)\), which are \((-1, 8)\) and \((-1, -4)\).

Step 3: Find the Co-vertices (Endpoints of the Minor Axis)

The co-vertices are located \(b\) units from the center along the minor axis. The minor axis is horizontal, so the endpoints are at \((-1 + 5, 2)\) and \((-1 - 5, 2)\), which give \((4, 2)\) and \((-6, 2)\).

Step 4: Calculate the Foci

The foci are \(c\) units from the center along the major axis. Compute \(c\) using \(c = \sqrt{a^{2} - b^{2}} = \sqrt{36 - 25} = \sqrt{11}\). Thus, the foci are at \((-1, 2 + \sqrt{11})\) and \((-1, 2 - \sqrt{11})\).

Step 5: Determine the Eccentricity

The eccentricity \(e\) is given by \(e = \frac{c}{a} = \frac{\sqrt{11}}{6}\). This measures how much the ellipse deviates from being a circle.

Key Concepts

Center of Ellipse

In an ellipse, the **center** plays a crucial role much like the center of a circle, though the shapes differ. The center is the midpoint between the major and minor axes.

For our given ellipse equation \( \frac{(x+1)^{2}}{25} + \frac{(y-2)^{2}}{36} = 1 \), the center \((h, k)\) can be easily identified once the equation is rewritten in its standard form, \( \frac{(x-h)^{2}}{b^{2}} + \frac{(y-k)^{2}}{a^{2}} = 1 \).

By comparing, it becomes apparent that \( h = -1 \) and \( k = 2 \). Therefore, the center of this ellipse is at \((-1, 2)\). This point is where the two axes cross and serves as a reference for locating other important points on the ellipse.

Foci of Ellipse

The **foci** (plural of focus) are two fixed points on the interior of an ellipse used in the formal definition and construction of the ellipse.

These points are crucial because, in an ellipse, the sum of the distances from any point on the ellipse to the two foci is constant. For the given ellipse \( \frac{(x+1)^{2}}{25} + \frac{(y-2)^{2}}{36} = 1 \), the length of the semi-major axis \(a\) is 6, and the semi-minor axis \(b\) is 5.

To find the foci, we calculate the distance \(c\) from the center to each focus using the formula \(c = \sqrt{a^{2} - b^{2}}\). Here, \(c = \sqrt{36 - 25} = \sqrt{11}\). Positioned along the vertical major axis, the foci are at \((-1, 2 + \sqrt{11})\) and \((-1, 2 - \sqrt{11})\). These points give the ellipse its elongated shape.

Ellipse Vertices

**Vertices** are the endpoints of the major axis, the longest diameter that stretches through the center of the ellipse.

In the ellipse described by \( \frac{(x+1)^{2}}{25} + \frac{(y-2)^{2}}{36} = 1 \), identifying the major axis as the longer vertical one means the vertices are aligned vertically. From the center at \((-1, 2)\), the vertices are found \(a\) units away, where \(a = 6\).

Hence, the vertices are located at:

• \((-1, 8)\)
• \((-1, -4)\)

These positions provide the tips of the ellipse along its lengthiest section, defining its size and space.

Eccentricity of Ellipse

The **eccentricity** of an ellipse is a measure of how much it deviates from a perfect circle, where a circle has an eccentricity of zero.

For our particular ellipse, eccentricity \(e\) can be computed using the formula \(e = \frac{c}{a}\), where \(c\) is the distance from the center to each focus and \(a\) is the semi-major axis length.

With \(c = \sqrt{11}\) and \(a = 6\), the eccentricity is:

• \(e = \frac{\sqrt{11}}{6}\)

This value indicates that the ellipse is noticeably elongated but not excessively so.