Problem 44 - Algebra and Trigonometry

Question

In Exercises \(37-50,\) graph each ellipse and give the location of its foci. $$ \frac{(x-4)^{2}}{4}+\frac{y^{2}}{25}=1 $$

Step-by-Step Solution

Answer

The graph of the ellipse has a center at (4,0), vertices at (9,0) and (-1,0), and foci at \((4\pm \sqrt{21}, 0)\).

Step 1: Identifying the Center, a, and b

Here, the denominator of the x-term gives \(a^{2} = 4\), thus \(a = 2\), and the denominator of the y-term is \(b^{2} = 25\), thus \(b = 5\). The center (h,k) of the ellipse is the point (4,0).

Step 2: Computing the Distance of the Foci to the Center

The distance c from the center to a focus is given by \(c = \sqrt{b^2 - a^2}\). Here, it turns out to be \(c = \sqrt{25 - 4} = \sqrt{21}\).

Step 3: Finding the Foci

The coordinates of the foci are \((h\pm c, k)\). Substituting the center point (4,0) and c, we get the foci to be \((4\pm \sqrt{21}, 0)\).

Step 4: Graphing the Ellipse

Next, graph the ellipse using the center, axes, and foci. The center is at (4,0), and the vertices are at (4±5,0)=(9,0) and (-1,0). Sketch an ellipse that goes through these vertices and has foci at \((4\pm \sqrt{21}, 0)\).

Key Concepts

Foci

In every ellipse, the foci are two special points located inside the ellipse. These points are not just random; they hold a geometric significance in defining the ellipse's shape. The sum of the distances from any point on the ellipse to each of the foci is always constant. This property distinguishes ellipses from other conic sections.
For our specific problem, the foci are located based on the formula used for calculating the distance from the center. Once we identify the center as (4,0) and calculate the distance factor or radius using \(c = \sqrt{b^2 - a^2}\), we can find the positions of these foci at \(4 \pm \sqrt{21}, 0\) in this exercise's coordinate plane.

Center of Ellipse

The center of an ellipse is crucial because it's the starting point for defining the whole shape.
In the given equation, \frac{(x-4)^{2}}{4} + \frac{y^{2}}{25} = 1\, we can determine the center by observing the \(h\) and \(k\) values from the equation. For an ellipse equation like \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1,\) the center is directly \( (h, k) \). Here, that places the center at (4,0).
This is the point from which we calculate distances to the vertices and foci, making it the key reference for plotting the entire ellipse.

Distance Formula for Foci

To locate the foci of an ellipse, we need to calculate the distance \(c\) from the center using the relation \c = \sqrt{b^2 - a^2}\. This formula comes from the property that for standard ellipses, the distance from each focus to any point on the curve sums up to the length of the major axis.
Remember, \(a\) and \(b\) represent the semi-major and semi-minor axes. For this equation, \(a^2\) corresponds to 4 and \(b^2\) to 25, which means \(a = 2\) and \(b = 5\). Plugging these into the formula gives us \(c=\sqrt{25-4}=\sqrt{21}\).
This mathematical step ensures we accurately determine the separation between the center and each focus, allowing us to pinpoint the foci correctly.

Graphing Ellipses

Graphing an ellipse correctly involves understanding its fundamental components such as the center, foci, and axes.
When graphing the ellipse \frac{(x-4)^{2}}{4} + \frac{y^{2}}{25} = 1\, start by marking its center at (4,0). Next, place the foci at \(4 \pm \sqrt{21}, 0\) based on the distance calculated.
The vertices of the ellipse are identified by \(h\pm b, k\), since here our ellipse is oriented vertically along the y-axis due to \(b > a\). This results in vertices at (4±5,0), which are (9,0) and (-1,0).
Ultimately, sketch a smooth, symmetric oval shape through these vertices with the known foci inside, completing an accurate graph. Remember, the sketch should reflect the symmetrical property of the ellipse and its constant distances from points on the ellipse to the foci.