Problem 31 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the center, vertices, foci, and eccentricity of the ellipse, and sketch its graph. Use a graphing utility to verify your graph. $$\frac{(x-4)^{2}}{16}+\frac{(y+1)^{2}}{25}=1$$

Step-by-Step Solution

Answer

The center of the ellipse is (4, -1), vertices are at (4,4) and (4,-6), foci are at (4,2) and (4,-4), and the eccentricity is 0.6.

Step 1: Determine the center

From the given equation \(\frac{(x-4)^{2}}{16}+\frac{(y+1)^{2}}{25}=1\), we can deduce that the center \((h, k)\) is located at the point (4, -1). This comes directly from comparing the equation with the standard form for an ellipse.

Step 2: find the vertices

The vertices lie on the major axis, at a distance of 'a' from the center. In this case, the major axis is vertical because b>a (25>16). Taking the square root of b^2 gives us the value for b length, i.e., 5. So, the vertices are \((h, k \pm b)\), i.e., \((4, -1 \pm 5)\), which gives us the vertices points (4,4) and (4,-6).

Step 3: Compute the foci

The foci are also located on the major axis, at a distance of 'c' from the center, where \(c=\sqrt{b^2-a^2}\). Here, \(c=\sqrt{25-16} =3\). Hence, the foci are at \((h, k \pm c)\), i.e., \((4, -1 \pm 3)\), which gives us the foci points (4,2) and (4,-4).

Step 4: Calculate the eccentricity

The eccentricity 'e' can be computed by the formula \(\frac{c}{b}\). Therefore, the eccentricity of this ellipse is 3/5 = 0.6.

Step 5: Sketch the graph

Plot the center, vertices and foci on a graph. Draw the major (vertical) and minor (horizontal) axes. Draw the ellipse so it intersects the vertices and is symmetric around the center (4,-1). Be sure to verify your graph with a graphing utility as well.

Key Concepts

Center of an Ellipse

The center of an ellipse is the midpoint around which the entire shape is symmetrically distributed. For the given ellipse equation: \[ \frac{(x-4)^{2}}{16} + \frac{(y+1)^{2}}{25} = 1 \] we determine the center by identifying the values that make the squared terms equal to zero. The standard form of ellipse equation is \[ \frac{(x-h)^{2}}{a^{2}} + \frac{(y-k)^{2}}{b^{2}} = 1 \] where

• \((h, k)\) represents the center coordinates.
• Comparing with our equation, the center of this ellipse is \((4, -1)\).

This point is where the ellipse is centered on the coordinate plane. Locating the center is essential for accurately plotting the ellipse and understanding its geometric properties.

Vertices of an Ellipse

Vertices are crucial points on an ellipse that determine its shape and size. They sit at the maximum extensions along the major axis – the longest diameter of the ellipse.For our equation given,

• The major axis is vertical because the denominator under the \(y\)-term (25) is larger than that under the \(x\)-term (16).
• Vertices are calculated using \((h, k \pm b)\).
• In this case, \(b = \sqrt{25} = 5\).
• Thus, the vertices are positioned at \((4, -1 \pm 5)\), giving us the coordinates:
  • (4, 4)
  • (4, -6)

These points tell us the full stretch of the ellipse vertically and are vital for the construction of the ellipse.

Foci of an Ellipse

The foci (singular: focus) are two special points inside an ellipse. Each point lies on the major axis and as you move along the curve of the ellipse, the sum of the distances from any point on the ellipse to each focus is constant.For our ellipse, we find the foci using

• The formula: \(c = \sqrt{b^2 - a^2}\). Here, \(a^2 = 16\) and \(b^2 = 25\).
• This results in \(c = \sqrt{25 - 16} = 3\).

Thus, the foci are located at \((h, k \pm c)\), resulting in points:

• (4, 2)
• (4, -4)

Understanding the foci provides insights into how 'stretched' or 'elongated' an ellipse is along its major axis.

Eccentricity of an Ellipse

Eccentricity measures how much an ellipse deviates from being circular.It's defined as \(e = \frac{c}{b}\) and ranges between 0 and 1:

• If \(e = 0\), the ellipse looks more like a circle.
• As \(e\) approaches 1, the ellipse becomes more elongated.

For the current ellipse,

• Using \(c = 3\) and \(b = 5\), we compute its eccentricity as \(rac{3}{5} = 0.6\)

This value indicates that the ellipse is moderately elongated, confirming the more prominent vertical stretch compared to a perfect circle.Knowing the eccentricity helps predict and visualize the overall shape of the ellipse.