Problem 16 - Precalculus with Calculus Previews

Question

Find the center, foci, vertices, endpoints of the minor axis, and eccentricity of the given ellipse. Graph the ellipse. $$ 6(x-2)^{2}+8 y^{2}=48 $$

Step-by-Step Solution

Answer

Center: (2, 0); Foci: (2±√2, 0); Vertices: (2±2√2, 0); Minor Axis: (2, ±√6); Eccentricity: 1/2

Step 1: Divide by 48

Start by dividing the entire equation by 48 to normalize it. This gives you: \[ \frac{6(x-2)^2}{48} + \frac{8y^2}{48} = \frac{48}{48} \]This simplifies to:\[ \frac{(x-2)^2}{8} + \frac{y^2}{6} = 1 \]

Step 2: Identify the center

The standard form for an ellipse is given by:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]From the equation \( \frac{(x-2)^2}{8} + \frac{y^2}{6} = 1 \), we see that \( h = 2 \) and \( k = 0 \). Thus, the center of the ellipse is \((2, 0)\).

Step 3: Identify the vertices

The vertices are found using the values of \(a\) and \(b\). Here, \(a^2 = 8\) and \(b^2 = 6\), so \( a = \sqrt{8} = 2\sqrt{2} \) and \( b = \sqrt{6} \).Since \(a > b\), the major axis is horizontal. The vertices are along the x-axis at a distance \(a\) from the center. Thus, the vertices are:\((h+a, k) = (2 + 2\sqrt{2}, 0)\) and \((h-a, k) = (2 - 2\sqrt{2}, 0)\).

Step 4: Identify the foci

The foci are found using \(c = \sqrt{a^2 - b^2}\). Compute:\[ c = \sqrt{8 - 6} = \sqrt{2} \]The foci are along the x-axis at a distance \(c\) from the center. Thus, the foci are:\((h+c, k) = (2 + \sqrt{2}, 0)\) and \((h-c, k) = (2 - \sqrt{2}, 0)\).

Step 5: Identify endpoints of the minor axis

The endpoints of the minor axis are determined by the value of \(b\), aligned vertically as the minor axis is vertical. Therefore, the endpoints are:\((h, k+b) = (2, \sqrt{6})\) and \((h, k-b) = (2, -\sqrt{6})\).

Step 6: Calculate the eccentricity

Eccentricity \(e\) of an ellipse is calculated using:\[ e = \frac{c}{a} = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2} \]

Step 7: Graph the ellipse

To graph the ellipse, plot the center at \((2, 0)\). Mark the vertices at \((2 + 2\sqrt{2}, 0)\) and \((2 - 2\sqrt{2}, 0)\), plot the foci at \((2 + \sqrt{2}, 0)\) and \((2 - \sqrt{2}, 0)\), and mark the endpoints of the minor axis at \((2, \sqrt{6})\) and \((2, -\sqrt{6})\). Sketch the ellipse by connecting these points smoothly.

Key Concepts

Equation of Ellipse

An ellipse is essentially a stretched circle, and its equation helps us understand its shape and orientation. In precalculus, the equation of an ellipse looks like this: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] Here,

• \( (h, k) \) is the center of the ellipse.
• \( a \) and \( b \) represent the distances from the center to the vertices along the x-axis and y-axis, respectively.

The exercise gave us the equation \( 6(x-2)^2 + 8y^2 = 48 \). We started by dividing all terms by 48 to adopt the standard form. After simplifying, the equation becomes \( \frac{(x-2)^2}{8} + \frac{y^2}{6} = 1 \). This tells us that the ellipse is centered at \((2, 0)\), with the horizontal radius or semi-major axis \( a = \sqrt{8} = 2\sqrt{2} \) and the vertical radius or semi-minor axis \( b = \sqrt{6} \).

Foci of Ellipse

The foci are two special points inside the ellipse where the sum of the distances from any point on the ellipse to the foci is constant. In the standard equation, these points are identified using the relationship \( c = \sqrt{a^2 - b^2} \). For our ellipse, we found \( a^2 = 8 \) and \( b^2 = 6 \), hence \( c = \sqrt{8 - 6} = \sqrt{2} \). Since the major axis is horizontal and the center is at \( (2, 0) \), the foci are located \( \sqrt{2} \) units away along the x-axis:

• First focus: \( (2 + \sqrt{2}, 0) \)
• Second focus: \( (2 - \sqrt{2}, 0) \)

These foci help in plotting the accurate shape of the ellipse on a graph.

Eccentricity of Ellipse

Eccentricity is a number that indicates how much an ellipse deviates from being a circle. A perfect circle has an eccentricity of 0, while an ellipse has an eccentricity between 0 and 1. In mathematical terms, eccentricity \( e \) is derived from the formula:\[ e = \frac{c}{a} \]In the exercise at hand, \( c = \sqrt{2} \) and \( a = 2\sqrt{2} \). Hence, the eccentricity is \( e = \frac{\sqrt{2}}{2\sqrt{2}} = \frac{1}{2} \). This value shows that the ellipse isn't very elongated; it's closer to being circular but still an ellipse. Understanding eccentricity helps visualize the ellipse's overall shape.

Vertices of Ellipse

Vertices are the key points at the end of the major axis of an ellipse. Knowing the vertices is important as they define the longest diameter of the ellipse. Given our standard form \( \frac{(x-2)^2}{8} + \frac{y^2}{6} = 1 \), we determined:

• \( a^2 = 8 \) resulting in \( a = 2\sqrt{2} \).
• Since \( a > b \), the major axis is along the x-axis.

Using the center \((2, 0)\) and \( a \), we locate the vertices:

• Left vertex: \( (2 - 2\sqrt{2}, 0) \)
• Right vertex: \( (2 + 2\sqrt{2}, 0) \)

These points guide us in sketching the ellipse by marking its most stretched components along the x-axis.