Problem 85 - Calculus Early Transcendentals

Question

Consider an ellipse to be the set of points in a plane whose distances from two fixed points have a constant sum 2 \(a .\) Derive the equation of an ellipse. Assume the two fixed points are on the \(x\) -axis equidistant from the origin.

Step-by-Step Solution

Answer

Question: Derive the equation of an ellipse with foci on the x-axis and equidistant from the origin. Answer: The equation of an ellipse with foci on the x-axis and equidistant from the origin is (x^2 / a^2 - 1) + y^2 / (a^2 - c^2) = 1, where a and c are constants.

Step 1: Introduction of variables and distance formula

Let the two fixed points be F1 (-c, 0) and F2 (c, 0), with the foci on the x-axis and equidistant from the origin. Let P (x, y) be any point on the ellipse. By definition, PF1 + PF2 = 2a, where PF1 and PF2 are the distances from P to the foci F1 and F2, respectively. To find these distances, we will use the distance formula: Distance(P, Q) = sqrt((Q_x - P_x)^2 + (Q_y - P_y)^2) for two points P and Q.

Step 2: Calculate PF1 and PF2 using the distance formula

Using the distance formula for PF1 and PF2, we get the following expressions: PF1 = sqrt((x - (-c))^2 + (y - 0)^2) = sqrt((x + c)^2 + y^2) PF2 = sqrt((x - c)^2 + (y - 0)^2) = sqrt((x - c)^2 + y^2)

Step 3: Apply the constant sum condition

Since PF1 + PF2 = 2a, we substitute the expressions for PF1 and PF2 to create an equation: sqrt((x + c)^2 + y^2) + sqrt((x - c)^2 + y^2) = 2a

Step 4: Solve the equation for the equation of an ellipse

To derive the equation of the ellipse, we need to eliminate the square roots. We start by isolating one of the square root terms: sqrt((x - c)^2 + y^2) = 2a - sqrt((x + c)^2 + y^2) Now, square both sides of the equation: ((x - c)^2 + y^2) = (2a - sqrt((x + c)^2 + y^2))^2 Expand the equation: (x^2 - 2cx + c^2 + y^2) = 4a^2 - 4a * sqrt((x + c)^2 + y^2) + (x + c)^2 + y^2 Combine like terms on both sides: x^2 + c^2 - 2cx + y^2 - (x^2 + c^2 + 2cx + y^2) = 4a^2 - 4a * sqrt((x + c)^2 + y^2) Simplify: -4cx = 4a^2 - 4a * sqrt((x + c)^2 + y^2) Now, divide both sides by -4: cx = -a^2 + a * sqrt((x + c)^2 + y^2) Square both sides again: c^2x^2 = a^4 - 2a^3 * sqrt((x + c)^2 + y^2) + a^2 * ((x + c)^2 + y^2) Isolate the square root term: 2a^3 * sqrt((x + c)^2 + y^2) = a^4 - a^2 * ((x + c)^2 + y^2) + c^2x^2 Divide by 2a^3: sqrt((x + c)^2 + y^2) = (a^4 - a^2 * ((x + c)^2 + y^2) + c^2x^2) / (2a^3) Now, square both sides one more time: ((x + c)^2 + y^2) = (a^4 - a^2 * ((x + c)^2 + y^2) + c^2x^2)^2 / (4a^6) Finally, simplify the equation: (x^2 / a^2 - 1) + y^2 / (a^2 - c^2) = 1 This is the equation of an ellipse with the two foci on the x-axis and equidistant from the origin.

Key Concepts

Distance Formula

When working with the coordinates of geometric shapes, the distance formula becomes a critical tool. It helps measure the distance between two points on the Cartesian plane.

Given two points, \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), the distance between them is expressed as:

• \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

This formula is derived from the Pythagorean theorem, which relates the sides of a right triangle. In the context of an ellipse, this formula helps us calculate the distances from any point on the ellipse to its two foci.

The process involves substituting specific points into the formula, as seen with points \( F_1(-c, 0) \) and \( F_2(c, 0) \). By doing so, one can ensure the correct calculation of these crucial distances.

Conic Sections

Conic sections are the curves obtained when a plane intersects with a double cone. They come in different shapes: circles, ellipses, parabolas, and hyperbolas. An ellipse is one of these fascinating figures, shaped like an elongated circle.

Conic sections arise in various natural phenomena and are essential in several scientific fields. An ellipse is unique because it is defined by the sum of distances from any point on its edge to two fixed points, which remains constant.

• This property is mathematically expressed as \( PF_1 + PF_2 = 2a \), where \( a \) is a constant value.
• In terms of graphing, ellipses have two axes: a longer major axis and a shorter minor axis.

Understanding these properties helps in deriving the general equation of an ellipse, which we achieve by combining the conic section principle with the distance formula.

Focus Points

Focus points (or foci) are key elements when discussing ellipses. They play a central role in the defining property that characterizes ellipses: the constant sum of distances.

For an ellipse, the two foci are symmetric and located on the major axis equidistant from the center. Their coordinates in a standard horizontal ellipse aligned along the x-axis can be represented as \((-c, 0)\) and \((c, 0)\).

• The distance between the foci is denoted as \(2c\).
• The relation \(c^2 = a^2 - b^2\) reflects the link between the ellipse parameters, where \(a\) and \(b\) are the semi-major and semi-minor axes, respectively.

These focus points are critical in applications such as satellite orbits and laser optics, where reflecting properties of ellipses exploited. Understanding how to compute these foci positions and their relations to the rest of the ellipse is pivotal for comprehending its structure and function.