Problem 39 - Precalculus

Question

Find the center, vertices, foci, and eccentricity of the ellipse. Then sketch the ellipse. $$\frac{(x+5)^{2}}{9 / 4}+(y-1)^{2}=1$$

Step-by-Step Solution

Answer

The center is \(-5,1\), vertices are \((-5 + 3/2, 1)\) and \((-5 - 3/2, 1)\), foci are at \((-5 + 1/2, 1)\) and \((-5 - 1/2, 1)\) and the eccentricity is \(\frac{2}{3}\).

Step 1: Identify Center

For the given elliptic equation \(\frac{(x+5)^{2}}{9 / 4}+(y-1)^{2}=1\), the center of the ellipse is given by the values \(-5\) and \(1\) for the variables x and y respectively. Hence the center will be \(-5,1\).

Step 2: Find the Vertices

The vertices of the ellipse are the points where the ellipse crosses the x and y-axis. The x-coordinates of the vertices can be found by setting \(y=0\) in the equation of the ellipse, and similarly finding the y-coordinates for the vertices by setting \(x=0\). Before this, it is important to notice that \(\frac{9}{4}\) is the square of \(\frac{3}{2}\), thus this means that the semi-major axis is \(\frac{3}{2}\). Hence, the vertices are \((h+k, v)\) and \((h-k, v)\), where \(h\) is the x-coordinate of the center, \(v\) is the y-coordinate, and \(k\) is the semi-major axis. This leads to the vertices being \((-5 + 3/2, 1)\) and \((-5 - 3/2, 1)\).

Step 3: Find the Foci

The foci of the ellipse will lie along the major axis and are distanced from the center by an amount of \(\sqrt{(semi-major\ axis)^2 - (semi-minor\ axis)^2}\). In this case the semi-minor axis is 1 (since coefficient of \(y^2\) is 1). Using the formula, the distance comes out to be \(\sqrt{(\frac{3}{2})^2 -1^2}\) which results in \(\frac{1}{2}\). Therefore the foci will be located at \((-5 + 1/2, 1)\) and \((-5 - 1/2, 1)\).

Step 4: Find the Eccentricity

The eccentricity \(e\) of the ellipse is given by the formula \(e = \sqrt{1- (semi-minor\ axis / semi-major\ axis)^2}\). Substituting the given semi-major axis and semi-minor axis in the formula gives \(e = \sqrt{1- (1 / \frac{3}{2})^2}\) which results in \(e = \frac{2}{3}\).

Step 5: Sketch the Ellipse

Finally, to sketch the ellipse on the plane, draw an x-y axis, and mark the center of the ellipse as identified in Step 1. Then mark the vertices located along the x-axis as identified in Step 2. Also, mark the two foci as identified in Step 3. Subsequently, draw a smooth curve to form the ellipse. Keep in mind the shape of the ellipse is defined by the values of its semi-major and semi-minor axes.

Key Concepts

Center of Ellipse

Let's start by exploring the center of an ellipse. The center is a significant point as it helps in defining the position of an ellipse on the graph. In general, the equation of an ellipse is written in the form:\[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \]where \(h\) and \(k\) are the x and y coordinates of the ellipse's center, respectively.

For our specific equation, \(\frac{(x+5)^2}{9/4} + (y-1)^2 = 1\), the center coordinates are determined by transforming the expressions \(x+5\), and \(y-1\). This results in:

• h = -5
• k = 1

Thus, the center of the ellipse is at the point \,(-5, 1). This point acts as the middle from which every feature of the ellipse, such as vertices and foci, is symmetrically oriented.

Vertices of Ellipse

Vertices are the points on an ellipse that lie at the extremities of its longest and shortest axes. To find these points for an ellipse, you need to evaluate the semi-major axis, which relates to the denominator of the x-related term if larger, or the y-related term otherwise.

For the equation \(\frac{(x+5)^2}{9/4} + (y-1)^2 = 1\), we identified that the semi-major axis length \(a\) is \(\frac{3}{2}\) because \[(\frac{9}{4} = (\frac{3}{2})^2)\]. The semi-minor axis derived from the y-term is \(b = 1\). Given that the x-term is larger, the semi-major axis is along the x-direction.

Using the center \,(-5, 1), we calculate vertices by adding and subtracting the semi-major axis length from the x-coordinate of the center:

• V1 = (-5 + \frac{3}{2}, 1) \Rightarrow (-3.5, 1)
• V2 = (-5 - \frac{3}{2}, 1) \Rightarrow (-6.5, 1)

These points mark where the ellipse reaches its maximum extent along the x-axis.

Foci of Ellipse

The foci are two special points located symmetrically along the major axis. They help to define the shape of an ellipse and are integral in its mathematical properties.

To find the foci, we use the formula that calculates their distance from the center:\[ c = \sqrt{a^2 - b^2} \]where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

For our ellipse, using the derived values:

• a = \frac{3}{2}
• b = 1

We find:\[ c = \sqrt{(\frac{3}{2})^2 - 1^2} = \frac{1}{2} \]

Therefore, the foci are at:

• F1 = (-5 + \frac{1}{2}, 1) \Rightarrow (-4.5, 1)
• F2 = (-5 - \frac{1}{2}, 1) \Rightarrow (-5.5, 1)

These points lie on the ellipse's major axis and contribute to its overall form.

Eccentricity of Ellipse

Eccentricity is a number that describes the elongation of an ellipse. It varies between 0 and 1, where 0 represents a perfect circle and values closer to 1 indicate a more stretched ellipse.

The formula to calculate the eccentricity \(e\) is:\[ e = \sqrt{1 - \left( \frac{b}{a} \right)^2} \]where \(a\) is the semi-major axis and \(b\) is the semi-minor axis.

In our example, from having \(a = \frac{3}{2}\) and \(b = 1\), the eccentricity becomes:\[ e = \sqrt{1 - \left( \frac{1}{\frac{3}{2}} \right)^2} = \frac{2}{3} \]

This value shows the degree of elongation: the ellipse stretches noticeably along its x-axis.