Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the given conic. Horizontal ellipse with center \((5,1)\), major diameter 10 , minor diameter 8

Step-by-Step Solution

Answer

The equation is \(\frac{(x-5)^2}{25} + \frac{(y-1)^2}{16} = 1\).

Step 1: Identify the center of the ellipse

The center of the ellipse is given as \((5,1)\). This will be used in the ellipse equation \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] where \((h, k)\) is the center.

Step 2: Determine semi-major and semi-minor axes

The given major diameter of the ellipse is 10, so the semi-major axis \(a\) is half of that, which is 5. The minor diameter is 8, thus the semi-minor axis \(b\) is 4.

Step 3: Determine orientation

The ellipse is oriented horizontally, which means that the x-term (\((x-h)^2\)) is divided by the larger denominator \(a^2\).

Step 4: Write the standard form of the ellipse equation

Using the identified parameters: center \((5,1)\), \(a=5\), and \(b=4\), substitute into the standard equation:\[ \frac{(x-5)^2}{5^2} + \frac{(y-1)^2}{4^2} = 1 \]Simplifying the values gives:\[ \frac{(x-5)^2}{25} + \frac{(y-1)^2}{16} = 1 \]

Key Concepts

Ellipse

An ellipse is a special type of conic section that appears as an elongated circle. Unlike a perfect circle, an ellipse has two distinct axes: a major axis, which is the longest diameter running through its center, and a minor axis, which is the shortest diameter. Picture an ellipse as a squished or stretched circle.

Ellipses have unique reflective properties, much like a whispering gallery; they reflect paths between their foci (the two fixed points inside the ellipse). In mathematics and geometry, we often describe an ellipse using an equation in its standard form, allowing us to see its center, axes, and orientation clearly.

Standard form equation

The standard form of an ellipse equation is crucial for understanding its geometric properties. The general form is \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\]Here,

• \((h,k)\) is the center of the ellipse.
• 'a' is the semi-major axis, dictating the ellipse's stretch along its x or y orientation.
• 'b' is the semi-minor axis, providing the stretch perpendicular to 'a'.

For a horizontal ellipse, 'a' is typically larger than 'b'. The larger value under the fraction determines which axis is the major axis.

This formula helps to quickly understand the positioning and size of the ellipse relative to the coordinate plane. With the center given, the formula can easily be solved to express the relationship between x and y coordinates across the ellipse.

Semi-major axis

The semi-major axis denotes half of the ellipse's longest diameter. In an equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] it's represented by 'a'. This axis indicates the distance from the ellipse's center to its outermost edge horizontally or vertically, depending on the orientation.

For a horizontal ellipse, the semi-major axis runs parallel to the x-axis. This property also determines the ellipse's width. In our original exercise, with a major diameter of 10, the semi-major axis, 'a', is computed by halving this measurement, giving us 5. Hence, the ellipse stretches further in the direction of this axis than the minor one.

Semi-minor axis

The semi-minor axis represents half of the ellipse's shortest diameter. If we consider the equation \[\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\] 'b' denotes the semi-minor axis.

It indicates the distance from the ellipse’s center to its boundary along the minor axis, which runs perpendicularly to the major one. This gives us insight into how extended the ellipse is along its narrower dimension.

In our exercise, with a minor diameter given as 8, the semi-minor axis 'b' is calculated by dividing this diameter by two, resulting in a length of 4. Thus, the ellipse is not as expansive in this direction, maintaining its elongated circular shape.