Problem 63 - College Algebra with Corequisite Support

Question

Find the equation of the ellipse that will just fit inside a box that is four times as wide as it is high. Express in terms of \(h,\) the height.

Step-by-Step Solution

Answer

The equation of the ellipse is \( \frac{x^2}{4h^2} + \frac{4y^2}{h^2} = 1 \).

Step 1: Understanding the Problem

We need to find the equation of an ellipse that fits perfectly inside a rectangle. The rectangle's width is four times its height. We will express everything in terms of the rectangle's height, denoted as \(h\).

Step 2: Determine the Dimensions of the Box

The height of the box is \(h\), and the width is four times the height. Therefore, the width is \(4h\). The center of the ellipse will be at the center of the box.

Step 3: Identify Semi-Major and Semi-Minor Axes

The semi-major axis of the ellipse, usually denoted by \(a\), will be half of the width of the box. Thus, \(a = \frac{4h}{2} = 2h\). The semi-minor axis, \(b\), will be half of the height of the box, so \(b = \frac{h}{2}\).

Step 4: Write the Equation of the Ellipse

The standard form of the equation for an ellipse centered at the origin is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). Substituting in the values for \(a\) and \(b\), we get: \[ \frac{x^2}{(2h)^2} + \frac{y^2}{\left(\frac{h}{2}\right)^2} = 1 \].

Step 5: Simplify the Equation

Calculate \((2h)^2 = 4h^2\) and \(\left(\frac{h}{2}\right)^2 = \frac{h^2}{4}\). The equation simplifies to: \[ \frac{x^2}{4h^2} + \frac{y^2}{\frac{h^2}{4}} = 1 \]. Finally, simplify further to obtain: \[ \frac{x^2}{4h^2} + \frac{4y^2}{h^2} = 1 \].

Key Concepts

Semi-Major Axis

In the context of an ellipse, the semi-major axis is the longest radius, extending from the center to the furthest edge of the ellipse. Think of it as half of the ellipse's longest diameter.
In our problem, the ellipse fits inside a rectangle where the width is four times the height. This condition determines the semi-major axis. Since the width is the longer side when the rectangle is oriented traditionally, the semi-major axis equals half of the rectangle's width.

• Width of rectangle: represented by the height as it is specified as four times the height, denoted as \( 4h \)
• Resulting semi-major axis: \( a = \frac{4h}{2} = 2h \)

The semi-major axis governs the primary stretch of the ellipse, ensuring it reaches from one end of the rectangle to the other along the longer side. This axis fundamentally defines the magnitude of the ellipse in the horizontal direction.

Semi-Minor Axis

While the semi-major axis is the longer radius of the ellipse, the semi-minor axis is the shorter one, extending from the center, perpendicular to the semi-major axis, to the closest edge of the ellipse. In this case, we focus on the height of the box to determine the semi-minor axis.
The height of the rectangle is given as \( h \). The semi-minor axis will therefore be half of this height:

• Height of rectangle: \( h \)
• Semi-minor axis: \( b = \frac{h}{2} \)

The semi-minor axis effectively establishes the ellipse's width in the vertical direction. It reflects how much the ellipse stretches in this direction inside the confines of the rectangle.

Standard Form of Ellipse

The equation for an ellipse in the standard form helps in expressing its shape and size mathematically. For an ellipse centered at the origin, the standard form is:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
Here, \( a \) represents the semi-major axis, and \( b \) signifies the semi-minor axis. For our specific problem, we have

• Semi-major axis, \( a = 2h \)
• Semi-minor axis, \( b = \frac{h}{2} \)

Substituting these into the standard form, we get:
\[ \frac{x^2}{(2h)^2} + \frac{y^2}{\left(\frac{h}{2}\right)^2} = 1 \]
Simplifying this yields
\[ \frac{x^2}{4h^2} + \frac{y^2}{\frac{h^2}{4}} = 1 \]
Further simplification leads to the final equation:
\[ \frac{x^2}{4h^2} + \frac{4y^2}{h^2} = 1 \]
This equation defines how the ellipse will perfectly fit within the given rectangle, expressing its horizontal and vertical spans.