Problem 15 - Precalculus Mathematics for Calculus

Question

Find the vertices, foci, and eccentricity of the ellipse. Determine the lengths of the major and minor axes, and sketch the graph. $$2 x^{2}+y^{2}=3$$

Step-by-Step Solution

Answer

The vertices are \((0, \pm \sqrt{3})\), foci are \((0, \pm \sqrt{3/2})\), eccentricity is approximately 0.71. Major axis length is \(2\sqrt{3}\) and minor is \(2\sqrt{3/2}\).

Step 1: Write the Equation in Standard Form

The given equation is \( 2x^2 + y^2 = 3 \). Divide every term by 3 to separate constants so that we can compare it with the general ellipse equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Doing this gives \( \frac{2x^2}{3} + \frac{y^2}{3} = 1 \), which then becomes \( \frac{x^2}{3/2} + \frac{y^2}{3} = 1 \).

Step 2: Identify the Values of a and b

From \( \frac{x^2}{3/2} + \frac{y^2}{3} = 1 \), we have \( a^2 = 3/2 \) and \( b^2 = 3 \). Therefore, \( a = \sqrt{3/2} \approx 1.22 \) and \( b = \sqrt{3} \approx 1.73 \). Here, since \( b > a \), the ellipse is vertical.

Step 3: Find the Vertices

For a vertical ellipse, the vertices are located at \((0, \pm b)\). Therefore, the vertices are \((0, \sqrt{3})\) and \((0, -\sqrt{3})\).

Step 4: Calculate the Foci

For a vertical ellipse, we use the formula \( c^2 = b^2 - a^2 \) to find the foci. Therefore, \( c^2 = 3 - \frac{3}{2} = \frac{3}{2} \). Calculating gives \( c = \sqrt{3/2} \approx 1.22 \). Thus, the foci are \((0, \pm \sqrt{3/2})\) or approximately \((0, \pm 1.22)\).

Step 5: Calculate the Eccentricity

The eccentricity \( e \) is given by \( e = \frac{c}{b} \). Thus, \( e = \frac{\sqrt{3/2}}{\sqrt{3}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.71 \).

Step 6: Determine Lengths of Major and Minor Axes

The length of the major axis (vertical) is \( 2b = 2\sqrt{3} \approx 3.46 \), and the length of the minor axis (horizontal) is \( 2a = 2\sqrt{3/2} \approx 2.44 \).

Step 7: Sketch the Graph

Draw an ellipse centered at the origin with vertical major axis of length approximately 3.46 and horizontal minor axis of length approximately 2.44. Plot the vertices at \((0, \sqrt{3})\) and \((0, -\sqrt{3})\), and plot the foci at \((0, \sqrt{3/2})\) and \((0, -\sqrt{3/2})\).

Key Concepts

Vertices

The vertices of an ellipse are the points where the ellipse intersects its axes. In a vertical ellipse, such as the one defined by the equation \( \frac{x^2}{3/2} + \frac{y^2}{3} = 1 \), the vertices are along the y-axis.

To find them, note the values given in the standard form: \( a^2 = \frac{3}{2} \) and \( b^2 = 3 \). This means our vertices are at \((0, \pm b)\).

• Calculate \( b \) as \( \sqrt{3} \approx 1.73 \).

Therefore, the vertices are located at \((0, \sqrt{3})\) and \((0, -\sqrt{3})\).

These are the furthest points on the ellipse from its center along the major axis.

Foci

The foci of an ellipse are two special points located along its major axis. They play an essential role in defining the shape of the ellipse, as each point on the ellipse is the sum of the distances to the foci.

For our vertical ellipse:

• We use the formula \( c^2 = b^2 - a^2 \) to determine the foci.
• Substituting, we find \( c^2 = 3 - \frac{3}{2} = \frac{3}{2} \).
• Hence, \( c = \sqrt{\frac{3}{2}} \approx 1.22 \).

The foci are thus found at \((0, \pm \sqrt{3/2})\) or approximately \((0, \pm 1.22)\).

These points provide the geometric focal points for the ellipse’s shape and are key in understanding its form.

Eccentricity

Eccentricity measures how "stretched" an ellipse is, with values ranging from 0 to 1.

For an ellipse, eccentricity \( e \) is calculated using:

• The formula \( e = \frac{c}{b} \).

Using our findings, \( c \) is \( \sqrt{3/2} \approx 1.22 \) and \( b = \sqrt{3} \approx 1.73 \). Substituting these:

\[ e = \frac{\sqrt{3/2}}{\sqrt{3}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \approx 0.71 \] The eccentricity tells us how much the ellipse deviates from being a circle. In this case, since \( e \) is 0.71, it shows a moderate degree of elongation.

A smaller eccentricity means the ellipse is more circular.

Major and Minor Axes

The major and minor axes of an ellipse are the longest and shortest diameters, respectively.

For our vertical ellipse:

• The major axis runs along the y-axis, with length calculated as \( 2b \) where \( b = \sqrt{3} \approx 1.73 \). Consequently, the major axis is \( 2 \times 1.73 \approx 3.46 \).
• The minor axis runs along the x-axis, with length determined as \( 2a \) with \( a = \sqrt{3/2} \approx 1.22 \). Thus, the minor axis measures \( 2 \times 1.22 \approx 2.44 \).

These axes give us a complete description of the ellipse’s scale and orientation, with the major axis showing its maximum length and the minor axis showing its minimum width.