Problem 15 - Algebra for College Students

Question

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ \frac{(x+1)^{2}}{9}+\frac{(y+2)^{2}}{16}=1 $$

Step-by-Step Solution

Answer

Center: (-1, -2); Vertices: (-1, 2) and (-1, -6); Endpoints: (2, -2) and (-4, -2); Foci: (-1, -2±√7).

Step 1: Identify the Center of the Ellipse

The ellipse equation is given in the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). Here, \(h = -1\) and \(k = -2\), so the center of the ellipse is at the point \((-1, -2)\).

Step 2: Determine the Lengths of the Major and Minor Axes

In the equation \( \frac{(x+1)^{2}}{9} + \frac{(y+2)^{2}}{16} = 1 \), \(a^2 = 9\) and \(b^2 = 16\). Since \(b^2 > a^2\), the major axis is vertical, and \(b = 4\) is the semi-major axis length, while \(a = 3\) is the semi-minor axis length.

Step 3: Find the Vertices of the Ellipse

The vertices are along the major axis. From the center at \((-1, -2)\), move \(b = 4\) units up and down along the y-axis. This gives the vertices at \((-1, -2 + 4) = (-1, 2)\) and \((-1, -2 - 4) = (-1, -6)\).

Step 4: Find the Endpoints of the Minor Axis

The endpoints of the minor axis are horizontal from the center along the x-axis. From \((-1, -2)\), move \(a = 3\) units left and right. This results in the endpoints \((-1 + 3, -2) = (2, -2)\) and \((-1 - 3, -2) = (-4, -2)\).

Step 5: Calculate the Foci of the Ellipse

The distance of the foci from the center is given by \(c = \sqrt{b^2 - a^2} = \sqrt{16 - 9} = \sqrt{7}\). The foci are along the major axis, so the coordinates of the foci are \((-1, -2 + \sqrt{7})\) and \((-1, -2 - \sqrt{7})\).

Step 6: Sketch the Ellipse

Plot the center, vertices, the endpoints of the minor axis, and the foci. Draw a smooth, symmetric ellipse around these points, ensuring that the major axis is vertical since \(b > a\).

Key Concepts

Ellipse Center

An ellipse's center is akin to its balancing point. For the equation \[\frac{(x+1)^{2}}{9} + \frac{(y+2)^{2}}{16} = 1,\]the center is given by the defaults from the general formula \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1.\]The center of an ellipse occurs at the point \((h, k)\). In this equation, observe that we have shifted from the origin. This is indicated by values within the squared terms:

• h is \(-1\), shifted from x
• k is \(-2\), shifted from y

Hence, the center of the ellipse is at \((-1, -2)\). The center is crucial because it serves as a reference for locating the vertices, endpoints of the minor axis, and foci.

Ellipse Vertices

Vertices of an ellipse represent the most extended points on the ellipse's major axis. Observing the given equation \[\frac{(x+1)^{2}}{9} + \frac{(y+2)^{2}}{16} = 1,\]we determine the length of the semi-major axis, \(b = 4\), which is derived from \(b^2 = 16\).

• The major axis is vertical because \(b > a\).
• To find the vertices, translate \(b = 4\) units up and down from the center point.
• This calculates to vertices at \((-1, 2)\) and \((-1, -6)\).

These vertices are crucial because they represent the farthest distance covered by the ellipse vertically. By establishing them, you lay important groundwork for understanding the ellipse's proportions and orientation.

Endpoints of Minor Axis

Minor axis endpoints define the shorter diameter of an ellipse. For the minor axis, examine the given ellipse equation:\[\frac{(x+1)^{2}}{9} + \frac{(y+2)^{2}}{16} = 1.\]Identifying the semi-minor axis length, \(a = 3\), stems from \(a^2 = 9\).

• The minor axis aligns horizontally since the semi-minor axis is associated with \(a\).
• From the center \((-1, -2)\), move horizontally \(3\) units. So, endpoints are situated at \((2, -2)\) and \((-4, -2)\).

These endpoints are pivotal as they highlight how wide the ellipse spans in the horizontal direction, further providing insight into the overall shape.

Ellipse Foci

Foci of an ellipse are two special internal points such that the sum of distances from them to the ellipse's boundary remains constant. Using the equation \[\frac{(x+1)^{2}}{9} + \frac{(y+2)^{2}}{16} = 1,\]we calculate foci:

• The formula \(c = \sqrt{b^2 - a^2}\) gives the foci's distance from the center along the major axis.
• Using \(b^2 = 16\) and \(a^2 = 9\), derive \(c = \sqrt{16 - 9} = \sqrt{7}\).
• The positions, \((-1, -2 + \sqrt{7})\) and \((-1, -2 - \sqrt{7})\), are vertically aligned with the ellipse's center because the major axis is vertical.

The foci are integral to understanding an ellipse's eccentricity, which influences how "stretched" the shape is compared to being a perfect circle.