Problem 31 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the foci of the ellipse $$ 16(x-1)^{2}+25(y+2)^{2}=400 $$

Step-by-Step Solution

Answer

The foci of the ellipse are at \((4, -2)\) and \((-2, -2)\).

Step 1: Identify the form of the equation

The given equation is \( 16(x-1)^2 + 25(y+2)^2 = 400 \). This can be compared to the standard form of an ellipse equation \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \). First, we'll rewrite the given equation in this standard form.

Step 2: Convert the equation to standard form

Divide through by 400 to convert the equation to standard form: \( \frac{16(x-1)^2}{400} + \frac{25(y+2)^2}{400} = 1 \). Simplify this to get \( \frac{(x-1)^2}{25} + \frac{(y+2)^2}{16} = 1 \). Now it is in standard form, where \( a^2 = 25 \) and \( b^2 = 16 \).

Step 3: Identify \(a\), \(b\), and \(c\)

In the standard ellipse equation, \(a^2\) corresponds to the larger denominator and \(b^2\) to the smaller if \(a > b\). Here we see \(a = 5\) and \(b = 4\). To find the foci, compute \(c\) using \(c^2 = a^2 - b^2\). Calculate \(c: c^2 = 25 - 16 = 9\), hence \(c = 3\).

Step 4: Determine the orientation of the ellipse

Since \(a^2 = 25\) is under \((x-1)^2\), the ellipse is oriented horizontally. This means the foci are along the x-axis, centered around \((h, k) = (1, -2)\).

Step 5: Find the coordinates of the foci

The foci of a horizontal ellipse are located at \((h \pm c, k)\). Substituting the values \(h = 1\), \(k = -2\), and \(c = 3\), we get the foci at \((1+3, -2)\) and \((1-3, -2)\). Thus, the coordinates are \((4, -2)\) and \((-2, -2)\).

Key Concepts

Ellipse Equation

Ellipses are fascinating shapes in geometry, distinguishable by their elongated circular form. The general equation of an ellipse is given in the form \( Ax^2 + By^2 + Cx + Dy + E = 0 \). This equation describes an ellipse's shape and properties. More precisely, the equation defines the relationship between the x and y coordinates on the plane. In many problems, such as finding foci, we start with a given equation that's not yet in the easiest form to manipulate. Here, it is essential to convert it into its standard form, which reveals more about the ellipse's characteristics. When you see an equation like \( 16(x-1)^2 + 25(y+2)^2 = 400 \), you need to recognize it as an ellipse equation."

Ellipse Standard Form

The standard form of an ellipse's equation is crucial for finding various properties of the ellipse, such as its foci. The standard form is:

• \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \) for a horizontal ellipse.
• \( \frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1 \) for a vertical ellipse.

Here, \((h, k)\) represents the center of the ellipse, \(a\) is the semi-major axis' length, and \(b\) is the semi-minor axis' length. By rewriting the equation \( 16(x-1)^2 + 25(y+2)^2 = 400 \) into its standard form: \( \frac{(x-1)^2}{25} + \frac{(y+2)^2}{16} = 1 \), vital characteristics become apparent. You know where to look for the lengths and direction of the axes based on the denominators and their respective variables."

Ellipse Orientation

Orientation is a key aspect of an ellipse, helping you understand how the ellipse is aligned on the coordinate plane. There are two primary orientations:

• Horizontal: Major axis runs parallel to the x-axis.
• Vertical: Major axis runs parallel to the y-axis.

Determining the orientation is straightforward once you have the standard form of the equation. Here, the equation \( \frac{(x-1)^2}{25} + \frac{(y+2)^2}{16} = 1 \) indicates a horizontal orientation because \( a^2 = 25 \) appears under the \( (x-1)^2 \) term, which is larger than \( b^2 \), showing it's the semi-major axis. This information tells us that the ellipse stretches wider horizontally rather than vertically."

Ellipse Calculation Steps

Solving problems involving ellipses often requires careful, systematic calculations. Follow these steps:

• Identify Equation Type: Know whether the equation fits the general pattern of an ellipse.
• Transform to Standard Form: Divide through by the constant to achieve the form \(1\), as done from \(16(x-1)^2 + 25(y+2)^2 = 400 \) to \(\frac{(x-1)^2}{25} + \frac{(y+2)^2}{16} = 1 \).
• Determine Orientation: Based on \(a^2\) and \(b^2\)'s relative sizes, infer whether the orientation is horizontal or vertical.
• Calculate Foci: Once you have \(a\) and \(b\), use \(c^2 = a^2 - b^2\) to find \(c\) and thus the foci, computed as \( (h \pm c, k) \).

These steps streamline the process of working through ellipse challenges, leading you to a solution efficiently and accurately."