Problem 101 - A Graphical Approach to Precalculus with Limits

Question

Equation of an Ellipse Use the definition of an ellipse to find an equation of an ellipse with foci \((3,0)\) and \((-3,0),\) where the sum of the distances from any point of the ellipse to the two foci is \(10 .\)

Step-by-Step Solution

Answer

The equation of the ellipse is \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \).

Step 1: Understand the Definition of an Ellipse

The definition of an ellipse is that it is the set of all points such that the sum of the distances from any point on the ellipse to the two foci is constant. This means that if you have a point \( (x, y) \) on the ellipse, then \(d_1 + d_2 = 10\), where \(d_1\) is the distance to \((3,0)\) and \(d_2\) is the distance to \((-3,0)\).

Step 2: Equation for Distances to Foci

The distance from a point \((x, y)\) to the focus \((3, 0)\) is given by \(d_1 = \sqrt{(x-3)^2 + y^2}\) and to the focus \((-3, 0)\) is \(d_2 = \sqrt{(x+3)^2 + y^2}\). So we have the equation \(\sqrt{(x-3)^2 + y^2} + \sqrt{(x+3)^2 + y^2} = 10\).

Step 3: Square Both Sides to Eliminate Square Roots

To simplify, square both sides to remove the square roots: \( \left(\sqrt{(x-3)^2 + y^2} + \sqrt{(x+3)^2 + y^2}\right)^2 = 10^2 \). This expands to \((x-3)^2 + y^2 + 2\sqrt{((x-3)^2 + y^2)((x+3)^2 + y^2)} + (x+3)^2 + y^2 = 100\).

Step 4: Simplify the Equation

Combine and simplify terms: \(2x^2 + 2y^2 + 18 + 2\sqrt{((x-3)^2 + y^2)((x+3)^2 + y^2)} = 100\). Thus, \(x^2 + y^2 + \sqrt{((x-3)^2 + y^2)((x+3)^2 + y^2)} = 50\).

Step 5: Determine the Standard Form of the Ellipse

Using algebraic manipulation to further simplify without the square root and recognizing symmetry with respect to the x-axis, an ellipse with a horizontal major axis is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). Here, the distance between the foci (2c = 6 implies c = 3), and 2a = 10, implies \(a = 5\). Use \(b^2 = a^2 - c^2\). So \(b^2 = 25 - 9 = 16\), giving \(b = 4\). Hence the equation is \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \).

Step 6: Write the Final Ellipse Equation

Combining all information, the equation of the ellipse is \( \frac{x^2}{25} + \frac{y^2}{16} = 1 \), representing an ellipse centered at the origin (0,0) with semi-major axis length 5 along the x-axis and semi-minor axis length 4 along the y-axis.

Key Concepts

Foci

An ellipse has two special points known as the foci (singular: focus). These points play a crucial role in defining the shape and size of the ellipse. The definition of an ellipse states that it is the set of all points for which the sum of the distances from two fixed points (the foci) is constant. In our exercise, the foci are

• (3, 0)
• (-3, 0)

Understanding the foci is essential because they determine how 'stretched' an ellipse is along its major axis. Despite their importance in shaping the ellipse, the foci are never on the ellipse itself but always lie inside it. They are part of the core elliptical geometry, which helps answer exercises and applications of ellipses.

Major Axis

The major axis of an ellipse is the longest diameter that runs through the center and both foci of the ellipse. It is a line that divides the ellipse into two equal halves and is known as the 'longest' part of the ellipse. For our ellipse with the given foci, the major axis is horizontal. This is because the foci themselves, and therefore the whole ellipse, are aligned along the x-axis. The length of the major axis is determined by the sum of the distances from any point on the ellipse to the foci. In our case, this sum is 10, making the major axis 10 units long. Since the major axis passes through the center at the origin (0,0), it stretches from -5 to +5 along the x-axis, which gives the ellipse its horizontal orientation.

Ellipse Equation

The equation of an ellipse represents the set of all points

• (x, y)

that lie on the ellipse. The standard form of the ellipse equation is \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \] where

• a is the semi-major axis length
• b is the semi-minor axis length

From the problem, we know:

• The distance between the foci is 6, giving us a value of c = 3 (half of the distance)
• The total length of the major axis is 10, so a = 5 (half of this length)
• To find the semi-minor axis, b, use the relationship \[ b^2 = a^2 - c^2 \] leading to \[ b^2 = 25 - 9 = 16 \] and thus, b = 4.

Plugging these into the standard form, we derive the ellipse equation: \[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \] This equation fully characterizes our ellipse located at the origin.

Distances to Foci

Distances to the foci are foundational in understanding ellipses. They help in deriving the equation of an ellipse and maintaining its geometric properties. For a given point on the ellipse, the sum of the distances to each focus is constant. This constancy is what gives the ellipse its shape. For our example, the sum of these distances from any point

• (x, y)

on the ellipse to the foci at

• (3, 0)
• (-3, 0)

is always 10. To find the distances, use the distance formula:

• \[ d_1 = \sqrt{(x-3)^2 + y^2} \]
• \[ d_2 = \sqrt{(x+3)^2 + y^2} \]

Calculating these distances and ensuring that

• \[ d_1 + d_2 = 10 \]

helps maintain the ellipse's roundness and its relation with the foci. Understanding these distances is key to comprehending how the equation and the physical shape of the ellipse connect.