Problem 3 - 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. $$ \frac{x^{2}}{4}+\frac{y^{2}}{9}=1 $$

Step-by-Step Solution

Answer

Vertices: \((0, \pm 3)\), Endpoints of Minor Axis: \((\pm 2, 0)\), Foci: \((0, \pm \sqrt{5})\).

Step 1: Identify the Form of the Ellipse Equation

The given equation is \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \). This fits the standard form of an ellipse equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a^2 < b^2 \).

Step 2: Determine 'a,' 'b,' and 'c'

The values of \( a^2 \) and \( b^2 \) are given as 4 and 9, respectively, meaning \( a = 2 \) and \( b = 3 \). The value of \( c \), which is the distance from the center to each focus, is given by \( c^2 = b^2 - a^2 \), so \( c^2 = 9 - 4 = 5 \), hence \( c = \sqrt{5} \).

Step 3: Find the Center of the Ellipse

Since the equation is centered at the origin, the center of the ellipse is \((0, 0)\).

Step 4: Determine the Vertices

The major axis is along the \(y\)-axis because \( b > a \). Thus, the vertices, located at the ends of the major axis, are at \( (0, \, \pm b) \) or \( (0, \, \pm 3) \).

Step 5: Determine the Endpoints of the Minor Axis

The minor axis is along the \(x\)-axis. The endpoints of the minor axis are located at \( (\pm a, \, 0) \) or \( (\pm 2, \, 0) \).

Step 6: Locate the Foci of the Ellipse

The foci of the ellipse are on the major axis, located at \( (0, \, \pm c) \) or \( (0, \, \pm \sqrt{5}) \).

Step 7: Sketch the Graph

To sketch the graph, draw an ellipse centered at the origin. The ellipse should extend 2 units along the \(x\)-axis, 3 units along the \(y\)-axis, and include the calculated vertices, endpoints of the minor axis, and foci.

Key Concepts

Vertices of an Ellipse

Understanding the vertices of an ellipse is essential as they mark the furthest points on the ellipse along its major axis. In our given equation \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \), we can spot that the major axis is oriented along the \( y \)-axis. This is because the denominator under \( y^2 \) (which is 9) is larger than the denominator under \( x^2 \) (which is 4), indicating a vertical major axis. From this equation, we know:

• \( b = 3 \) since \( b^2 = 9 \)
• \( a = 2 \) since \( a^2 = 4 \)

The vertices are located at the endpoints of the major axis. When the ellipse is centered at the origin \((0, 0)\), the vertices appear at \( (0, \pm b) \). In this case, they are at the coordinates \( (0, 3) \) and \( (0, -3) \). These are the highest and lowest points of the ellipse, respectively.

Foci of an Ellipse

The foci of an ellipse are two particular points located along the major axis. They are crucial as they define the shape and elongation of the ellipse. For the equation \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \), we need to find the value of \( c \), the distance from the center to each focus, given by the formula \( c^2 = b^2 - a^2 \).Here's the calculation:

• \( b^2 = 9 \) and \( a^2 = 4 \)
• Therefore, \( c^2 = 9 - 4 = 5 \)
• \( c = \sqrt{5} \)

That means the foci are located at \( (0, \pm \sqrt{5}) \), since the major axis is vertical. These points are crucial in determining the "flattening" of the ellipse and indicate how stretched it is compared to a circle. Remember, the foci always lie within the ellipse!

Graph of an Ellipse

Graphing an ellipse might sound complicated, but with the vertices, endpoints of the minor axis, and foci, it becomes simpler. The given ellipse \( \frac{x^2}{4} + \frac{y^2}{9} = 1 \) is centered at the origin \((0, 0)\).To sketch it accurately:

• Plot the vertices at \((0, 3)\) and \((0, -3)\) on the \( y \)-axis.
• Mark the endpoints of the minor axis at \((2, 0)\) and \((-2, 0)\) on the \( x \)-axis.
• Locate the foci at \((0, \sqrt{5})\) and \((0, -\sqrt{5})\) on the \( y \)-axis.

Using these points, sketch the elongated shape of the ellipse with a more extended length along the \( y \)-axis. The ellipse should touch the vertices and the minor axis endpoints, encircling the foci within its interior. The result is a neat, symmetric oval shape characteristic of ellipses.