Problem 38 - College Algebra

Question

For the following exercises, find the foci for the given ellipses. $$ \frac{x^{2}}{2}+\frac{(y+1)^{2}}{5}=1 $$

Step-by-Step Solution

Answer

The foci are at \((0, -1 + \sqrt{3})\) and \((0, -1 - \sqrt{3})\).

Step 1: Identify the Standard Form

The given ellipse equation is \( \frac{x^{2}}{2} + \frac{(y+1)^{2}}{5} = 1 \). This is of the form \( \frac{x^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), which is a vertically oriented ellipse because \( b^2 > a^2 \). Here, \( a^2 = 2 \) and \( b^2 = 5 \).

Step 2: Calculate Semi-Major and Semi-Minor Axes

Identify the semi-major axis \( b \) and semi-minor axis \( a \) from \( a^2 = 2 \) and \( b^2 = 5 \). Therefore, \( a = \sqrt{2} \) and \( b = \sqrt{5} \).

Step 3: Determine the Distance to the Foci

For an ellipse, the distance \( c \) from the center to each focus is determined by the equation \( c^2 = b^2 - a^2 \). Substitute the values: \( c^2 = 5 - 2 = 3 \), so \( c = \sqrt{3} \).

Step 4: Find the Center of the Ellipse

The center of the ellipse is obtained from rewriting the term \( (y+1)^2 \) as \( (y - (-1))^2 \). Therefore, the center is at \( (0, -1) \).

Step 5: Locate the Foci

Since the ellipse is vertical, the foci will be located along the vertical axis at a distance \( c \) from the center. Therefore, the foci are at \( (0, -1 + \sqrt{3}) \) and \( (0, -1 - \sqrt{3}) \).

Key Concepts

Ellipse Equation

An ellipse is a closed curve, resembling a stretched circle. The standard form equation for an ellipse centered at \( (h, k) \) is \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).
Here, \( a \) and \( b \) are the lengths of the semi-axes. This form gives crucial insights into the dimensions and orientation of the ellipse.
Understanding which term has the larger denominator helps us identify the orientation:

• If \( a^2 > b^2 \), the ellipse is horizontal, stretched along the x-axis.
• If \( b^2 > a^2 \), the ellipse is vertical, stretched along the y-axis.

For example, in the given equation \( \frac{x^{2}}{2} + \frac{(y+1)^{2}}{5} = 1 \), it fits the form of \( \frac{x^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), denoting a vertical ellipse since \( b^2 > a^2 \).
This helps us determine other properties like the lengths of the semi-major and semi-minor axes, as well as the placement of the ellipse's center and foci.

Semi-Major Axis

The semi-major axis of an ellipse is the longest radius extending from the center to the boundary of the ellipse. It’s denoted by \( b \) for a vertically oriented ellipse, where \( b^2 \) is the larger value in the standard form equation.
In the equation \( \frac{x^{2}}{2} + \frac{(y+1)^{2}}{5} = 1 \), \( b^2 = 5 \). Calculating the semi-major axis, we find \( b = \sqrt{5} \).
A few important points to remember:

• The semi-major axis dictates the longest reach of the ellipse along its primary orientation.
• Its direction and length provide insights into the overall shape and size of the ellipse.
• In vertically oriented ellipses, this axis runs parallel to the y-axis.

The semi-major axis is crucial in calculating the distance of the foci from the center, which is another defining feature of any ellipse.

Semi-Minor Axis

The semi-minor axis represents the shortest radius of the ellipse, extending from its center to the ellipse's edge. In a vertically oriented ellipse like ours, it is given by \( a \). In the standard form, it appears as the smaller value under the x-term.
From the given equation \( \frac{x^{2}}{2} + \frac{(y+1)^{2}}{5} = 1 \), we have \( a^2 = 2 \), resulting in \( a = \sqrt{2} \).Some key aspects include:

• The semi-minor axis is always perpendicular to the semi-major axis.
• Its measurement helps define the compactness or "narrowness" of the ellipse.
• For this ellipse, it aligns horizontally across the x-axis.

Understanding both axes helps in creating a complete picture of the ellipse's dimensions, fully understanding its structure and allowing for precise graph plotting.

Distance to Foci

The foci of an ellipse are two points located along its major axis that help in defining its shape. The distance from the center of the ellipse to each focus is crucial for its mathematical representation. This distance is denoted as \( c \) and can be calculated using the relationship \( c^2 = b^2 - a^2 \).
Applying this to our example, given \( b^2 = 5 \) and \( a^2 = 2 \), then \( c^2 = 5 - 2 = 3 \), which gives us \( c = \sqrt{3} \).Key points include:

• The foci are positioned at equal distance \( c \), from the center along the major axis.
• For a vertically oriented ellipse, they lie vertically at coordinates \( (0, -1 + \sqrt{3}) \) and \( (0, -1 - \sqrt{3}) \).

Locating the foci assists in plotting the ellipse accurately and uncovering its geometric properties. This calculation reinforces how the ellipse's stretched and compact aspects balance one another in its symmetrical structure.