Problem 41 - Thomas Calculus

Question

You may wish to review Section 1.2 before solving Exercises \(39-56\) The ellipse \(\left(x^{2} / 16\right)+\left(y^{2} / 9\right)=1\) is shifted 4 units to the right and 3 units up to generate the ellipse $$ \frac{(x-4)^{2}}{16}+\frac{(y-3)^{2}}{9}=1 $$ $$\begin{array}{l}{\text { a. Find the foci, vertices, and center of the new ellipse. }} \\ {\text { b. Plot the new foci, vertices, and center, and sketch in the new }} \\ {\text { ellipse. }}\end{array}$$

Step-by-Step Solution

Answer

The center of the ellipse is \((4,3)\), vertices are \((0,3)\) and \((8,3)\), and foci are \((4+\sqrt{7},3)\) and \((4-\sqrt{7},3)\).

Step 1: Understand the Standard Equation of an Ellipse

The standard equation of an ellipse is \(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\), where \((h,k)\) is the center, \(a\) is the semi-major axis, and \(b\) is the semi-minor axis. In the given equation \(\frac{(x-4)^{2}}{16}+\frac{(y-3)^{2}}{9}=1\), the center \((h,k)\) is \((4,3)\), \(a^2 = 16\), and \(b^2 = 9\). Since \(a^2 > b^2\), this is a horizontal ellipse.

Step 2: Calculate the Lengths of Axes

For the semi-major axis, calculate \(a = \sqrt{16} = 4\). For the semi-minor axis, calculate \(b = \sqrt{9} = 3\). This means the ellipse extends 4 units horizontally from the center and 3 units vertically.

Step 3: Find the Vertices

Vertices for a horizontal ellipse are located at \((h \pm a, k)\). Substitute \(a = 4\) and \(h = 4\): Vertices are \((4+4, 3) = (8,3)\) and \((4-4, 3) = (0,3)\).

Step 4: Find the Foci

The foci \((c)\) can be found using \(c=\sqrt{a^2-b^2}\). Calculate \(c = \sqrt{16 - 9} = \sqrt{7}\). The foci are located at \((h \pm c, k)\), which gives \((4+\sqrt{7}, 3)\) and \((4-\sqrt{7}, 3)\).

Step 5: Plot and Sketch the Ellipse

Plot the center at \((4, 3)\), the vertices at \((0, 3)\) and \((8, 3)\), and the foci at \((4+\sqrt{7}, 3)\) and \((4-\sqrt{7}, 3)\). Draw the ellipse using the vertices to guide its shape.

Key Concepts

Conic Sections

Conic sections are the curves obtained by intersecting a cone with a plane. They include circles, ellipses, parabolas, and hyperbolas. The shape and type of the conic section depend on the angle at which the plane intersects the cone.
In our exercise, we are dealing with an ellipse, which is a type of conic section. Ellipses are formed when the plane cuts through the cone at an angle, creating an enclosed, oval shape. This particular ellipse is represented in the standard form \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] where \((h, k)\) is the center of the ellipse.
Ellipses have two axes: the major axis, which is the longest diameter, and the minor axis, the shortest diameter. By analyzing these axes and their respective lengths, we gain insights into the ellipse's orientation and dimensions. Understanding these forms and their geometric significance is crucial for identifying the key elements such as center, vertices, and foci in conic sections.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, enables the study of geometry using a coordinate system. This approach is vital for precisely locating points and curves within a geometric plane, allowing algebraic representations of geometric figures like ellipses.
In this exercise, the ellipse is shifted to a new location on the coordinate plane, demonstrating the power of coordinate transformations. The original ellipse equation is expressed as \[\frac{(x^2)}{16} + \frac{(y^2)}{9} = 1\] which is then shifted 4 units right and 3 units up to become \[\frac{(x-4)^2}{16} + \frac{(y-3)^2}{9} = 1\].
Such transformations involve altering the coordinates of the center, vertices, and other significant points. Finding the new center at \((4, 3)\) and vertices at \((8, 3)\) and \((0, 3)\) illustrates how coordinate geometry enables precise shifts and detailed mapping in the plane.

Analytic Geometry

Analytic geometry, which uses algebra to study geometric properties, allows us to find key features of the ellipse mathematically. We use equations representing the curves to calculate properties like foci, vertices, and axes lengths.
For an ellipse, the semi-major axis \(a\) and semi-minor axis \(b\) define its dimensions. In the standard form equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\], \(a\) and \(b\) are essential for calculations. In our example, \(a^2 = 16\) and \(b^2 = 9\) lead us to compute \(a = 4\) and \(b = 3\). The semi-major axis determines the direction of the ellipse's elongation, making identifying whether the ellipse is horizontal or vertical straightforward.
Using the relationship \(c = \sqrt{a^2 - b^2}\), we find the distance of the foci from the center. This distance helps in locating the foci along the major axis, further deepening our understanding of the ellipse's geometric configuration.