Problem 55 - Intermediate Algebra

Question

The graph of equation is an ellipse. Determine which distance is longer, the distance between the \(x\)-intercepts or the distance between the \(y\) -intercepts. How much longer? \(\frac{x^{2}}{16}+\frac{y^{2}}{25}=1\)

Step-by-Step Solution

Answer

The distance between the y-intercepts is 2 units longer than the x-intercepts.

Step 1: Identify the Form of the Ellipse

The given equation is \(\frac{x^2}{16} + \frac{y^2}{25} = 1\), which is of the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) where \(a^2 = 16\) and \(b^2 = 25\). This indicates it is a vertical ellipse with a longer vertical axis (\(b > a\)).

Step 2: Calculate the x-Intercepts

Set \(y = 0\) in the equation to find the \(x\)-intercepts: \(\frac{x^2}{16} = 1\). Thus, \(x^2 = 16\), so \(x = \pm 4\). The \(x\)-intercepts are \((4, 0)\) and \((-4, 0)\).

Step 3: Determine the Distance Between x-Intercepts

The distance between the \(x\)-intercepts \((4, 0)\) and \((-4, 0)\) is calculated as \(4 - (-4) = 8\).

Step 4: Calculate the y-Intercepts

Set \(x = 0\) in the equation to find the \(y\)-intercepts: \(\frac{y^2}{25} = 1\). Thus, \(y^2 = 25\), so \(y = \pm 5\). The \(y\)-intercepts are \((0, 5)\) and \((0, -5)\).

Step 5: Determine the Distance Between y-Intercepts

The distance between the \(y\)-intercepts \((0, 5)\) and \((0, -5)\) is calculated as \(5 - (-5) = 10\).

Step 6: Compare Intercept Distances

Compare the two distances: the distance between \(x\)-intercepts is 8, and the distance between \(y\)-intercepts is 10.

Step 7: Calculate How Much Longer

The distance between the \(y\)-intercepts is longer. The difference is \(10 - 8 = 2\).

Key Concepts

x-Intercepts of an Ellipse

The concept of x-intercepts is crucial when analyzing the properties of an ellipse. These are the points where the ellipse crosses the x-axis. To find these points in an ellipse equation, you set the y-coordinate to zero and solve for x.
In the given problem, the ellipse equation is \(\frac{x^2}{16} + \frac{y^2}{25} = 1\). To find the x-intercepts:

• Set \(y = 0\) in the equation\(\Rightarrow \frac{x^2}{16} = 1\).
• Solve for \(x\), yielding \(x^2 = 16\).
• The solutions \(x = \pm 4\) provide the x-intercepts as the points \((4, 0)\) and \((-4, 0)\).

Recognizing these points helps in understanding the symmetry and orientation of the ellipse on the coordinate axis.

y-Intercepts of an Ellipse

Just like x-intercepts, y-intercepts provide insight into where the ellipse crosses the y-axis. Here, you set the x-coordinate to zero and solve for y to find the y-intercepts.
Using our equation \(\frac{x^2}{16} + \frac{y^2}{25} = 1\), the process is as follows:

• Set \(x = 0\) in the equation \(\Rightarrow \frac{y^2}{25} = 1\).
• Solve for \(y\), which results in \(y^2 = 25\).
• This yields \(y = \pm 5\), providing y-intercepts at \((0, 5)\) and \((0, -5)\).

These y-intercepts show the extent of the ellipse along the y-axis, confirming its vertical orientation.

Distance Calculation Between Intercepts

Calculating distances between intercepts allows us to quantitatively compare them. The simple subtraction method gives the linear distance between two intercepts.
For the x-intercepts \((4, 0)\) and \((-4, 0)\):

• The distance is \(4 - (-4) = 8\) units

For the y-intercepts \((0, 5)\) and \((0, -5)\):

• The distance is \(5 - (-5) = 10\) units

Understanding these calculations not only helps with analytical geometry but also with comprehending the physical span of the ellipse across its axes.

Vertical Ellipse Orientation

The orientation of an ellipse describes its longer axis, which can either be horizontal or vertical. In this problem, the ellipse is vertical because the value of \(b^2\) (25) is greater than \(a^2\) (16).
This affects the intercept distances, with the y-axis being the longer axis. A vertical ellipse has:

• A major axis along the y-axis.
• The y-intercepts more spread out than the x-intercepts.

By knowing the orientation, one can easily predict the relative distances between intercepts, as seen when the y-intercept distance is 10 units and the x-intercept distance is 8 units. Such understanding is pivotal in graphing and analyzing elliptical equations.