Problem 1 - College Algebra

Question

Graph each ellipse and locate the foci. $$ \frac{x^{2}}{16}+\frac{y^{2}}{4}=1 $$

Step-by-Step Solution

Answer

The ellipse has its vertices at (±4, 0), co-vertices at (0, ±2), and foci at (±2√3, 0).

Step 1: Identify ellipse parameters

In the given equation \( \frac{x^{2}}{16}+\frac{y^{2}}{4}=1 \), we can see that \( a^{2} = 16 \) and \( b^{2} = 4 \). So, \( a = 4 \) and \( b = 2 \)

Step 2: Calculate distance to the foci

Recall that \( c = \sqrt{a^{2}-b^{2}} \). So, \( c = \sqrt{4^2 - 2^2} = \sqrt{12} \) which simplifies to \( c = 2\sqrt{3} \)

Step 3: Sketch the ellipse

The given equation represents an ellipse with a horizontal major axis. The vertices are at (±4, 0), the co-vertices are at (0, ±2), and the foci are at (±2√3, 0).

Step 4: Locate the foci

The foci are found using the calculated value of \( c \) and are located at \( (c, 0) \) and \( (-c, 0) \) or \( (±2\sqrt{3}, 0) \)

Key Concepts

Graphing Ellipses

When it comes to graphing ellipses, you start by examining its equation and understanding the parameters involved.
The general form of an ellipse equation is either \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} + \frac{x^2}{b^2} = 1 \). In our specific case, we have \( \frac{x^{2}}{16} + \frac{y^{2}}{4} = 1 \). This tells us several things right away.

• The center of the ellipse is at the origin (0, 0) since there are no \(x\) or \(y\) terms outside of the squares.
• This ellipse lies horizontally because the larger denominator is under the \(x^2\) term (\( 16 > 4 \)).

By plotting the vertices at (±4, 0) and the co-vertices at (0, ±2), you can sketch the ellipse while ensuring accuracy in symmetry.

Ellipse Equation

An ellipse equation in its standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) provides a foundation in understanding its shape and orientation.
Here, variables \(a\) and \(b\) are crucial.

• \(a\) represents the semi-major axis length. For \( \frac{x^2}{16} \), 16 is \( a^2 \), thus \( a = 4 \).
• \(b\) is the semi-minor axis length. For \( \frac{y^2}{4} \), 4 is \( b^2 \), which gives \( b = 2 \).

This standard form indicates the ellipse is horizontally oriented due to \( a > b \).
The larger axis, referred to as the major axis, dictates the ellipse's stretch, while the smaller axis, the minor axis, runs perpendicular.

Foci of an Ellipse

The foci are two specific points within an ellipse that help define its shape.
They bear an essential role in understanding the geometric structure, located along the major axis.To find the distance to each focus (denoted as \( c \)), use the formula \( c = \sqrt{a^2 - b^2} \).
From our equation, we have:

• \( a^2 = 16 \) and \( b^2 = 4 \)
• Therefore, \( c = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \)

This calculation means the foci are positioned symmetrically at (±2√3, 0) along the x-axis, enhancing the ellipse's geometrical significance and symmetry.

Horizontal Major Axis

An ellipse with a horizontal major axis indicates that it stretches more along the x-axis compared to the y-axis.
This orientation arises because \( a > b \) in its equation. In this specific example, \( a = 4 \) and \( b = 2 \).Key characteristics include:

• The vertices of the ellipse, marking the farthest stretch along the x-axis, are located at (±4, 0).
• The foci, which add to the ellipse's unique definition, sit within this axis at (±2√3, 0).
• The co-vertices, representing the stretch along the y-axis, reside at (0, ±2).

Understanding the orientation of the major axis aids in visualizing the ellipse structure and allows for accurate sketching and analysis.