Problem 34 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 31-34: Find an equation for the set of points in an \(x y\)-plane that are equidistant from the point \(P\) and the line \(l\). $$ P(5,-2) ; \quad l: y=4 $$

Step-by-Step Solution

Answer

The equation is \( x^2 - 10x + 8y + 29 = 0 \).

Step 1: Understand the Problem

We are asked to find the equation of the set of points equidistant from a point \( P(5, -2) \) and a horizontal line \( l: y=4 \) in the \(xy\)-plane.

Step 2: Equation Format

The set of equidistant points forms a parabola. The geometric definition of a parabola is the set of points equidistant from a point (focus) and a line (directrix). Here, \( P(5, -2) \) is the focus, and \( y = 4 \) is the directrix.

Step 3: Distance from Point to Line

To find the equation, we need to equate the distance from any point \((x, y)\) on the parabola to \( P(5, -2) \) and the vertical distance from \( (x, y) \) to the line \( y = 4 \).

Step 4: Distance to Point P

The distance from a point \((x, y)\) to the point \( P(5, -2) \) is given by the formula: \[ \sqrt{(x - 5)^2 + (y + 2)^2} \]

Step 5: Distance to the Line

The distance from the point \((x, y)\) to the line \( y = 4 \) is the absolute difference in their y-values: \[ |y - 4| \]

Step 6: Equating Distances

For the set of points to be equidistant from \( P \) and the line, we set the two distances equal: \[ \sqrt{(x - 5)^2 + (y + 2)^2} = |y - 4| \]

Step 7: Remove Absolute Value

Since the distance to the line as an absolute value must be non-negative, we can use \[ \sqrt{(x - 5)^2 + (y + 2)^2} = y - 4 \] for simplicity, as we assume points are between or below both references when solving initially.

Step 8: Square Both Sides

Square both sides of the equation to eliminate the square root:\[ (x - 5)^2 + (y + 2)^2 = (y - 4)^2 \]

Step 9: Simplify Equation

Expanding both sides:- Left side: \[ (x - 5)^2 + (y + 2)^2 = x^2 - 10x + 25 + y^2 + 4y + 4 \]- Right side: \[ (y - 4)^2 = y^2 - 8y + 16 \]Equating them:\[ x^2 - 10x + 25 + y^2 + 4y + 4 = y^2 - 8y + 16 \]

Step 10: Rearrange Terms

Combine like terms on both sides to simplify:\[ x^2 - 10x + 25 + 4y + 4 = -8y + 16 + y^2 \] After simplifying by canceling \( y^2 \) on both sides,\[ x^2 - 10x + 8y + 29 = 0 \]

Step 11: Final Equation

The equation of the parabola is:\[ x^2 - 10x + 8y + 29 = 0 \]. This equation represents the locus of points equidistant from the point \( P(5,-2) \) and the line \( y=4 \).

Key Concepts

Equidistant Points

In geometry, equidistant points refer to a set of points that have the same distance from two or more objects. When dealing with parabolas, these objects are typically a point known as the focus and a line called the directrix. The unique feature of a parabola is that it consists of all points equidistant from both the focus and the directrix. This property can be used to derive the parabola's equation. For example, considering any point \((x, y)\) on a parabola, the distance to a fixed point, P, can be equated to the vertical distance to a fixed line, L. With this understanding, we can visually and mathematically establish the parabola's characteristics from its geometric properties. By using equations for distances, we can set these distances equal, leading to the parabolic form.

Focus

The focus of a parabola is a fixed point used in its geometric definition and equation formation. The position of the focus influences the curvature and orientation of the parabola. In the given problem, the focus is at the point \(P(5, -2)\).

• Being a central point in the parabola's structure, the focus determines one end of the equidistance condition.
• Its distance to a point on the parabola gives half of the required condition for forming the parabola.
The equation for the distance from any point \((x, y)\) on the parabola to the focus \(P\) can be calculated using the distance formula: \[\sqrt{(x - 5)^2 + (y + 2)^2}\]This geometric role makes the focus an integral part of understanding and solving problems involving parabolas.

Directrix

A directrix is a straight line that, along with the focus, defines a parabola. It serves as a reference line from which distances to points on the parabola are calculated. In this exercise, the directrix is the horizontal line \( y = 4 \).

• The directrix is crucial as it, together with the focus, governs the overall shape of the parabola.
• The distance from any point \((x, y)\) on the parabola to this line must equal the distance to the focus.

This distance is simpler to calculate than the distance to the focus, being the absolute difference in y-values: \[|y - 4|\]Understanding the directrix's role helps clarify how a parabola is symmetrically balanced between the focus and the line, ensuring its geometric properties.

Coordinate Geometry

Coordinate geometry, or analytic geometry, is a mathematical discipline combining algebra and geometry. It uses a coordinate system to investigate geometric shapes' properties and relationships. This approach is particularly useful for solving problems involving parabolas by translating geometric definitions into algebraic equations.

• Graphing points, lines, and curves on the coordinate plane allows visual and algebraic understanding of their properties.
• Coordinate geometry provides tools to derive equations like that of a parabola from geometric conditions.

In coordinate geometry, a parabola such as in the exercise is defined by the locus of points equidistant from the focus and the directrix. Using the distance formulas:- From a point to a focus: \( \sqrt{(x - 5)^2 + (y + 2)^2} \)- From a point to a horizontal line: \( |y - 4| \)We can form the equation of the parabola by equating these expressions, a method that makes use of coordinate geometry's capacity to blend geometry and algebra seamlessly.