Problem 15 - The Calculus with Analytic Geometry

Question

Find an equation that must be satisfied by the coordinates of any point whose distance from the point \((5,3)\) is always two units greater than its distance from the point \((-4,-2)\).

Step-by-Step Solution

Answer

Simplify: \((x - 5)^2 + (y - 3)^2 = (x + 4)^2 + (y + 2)^2 + 4 \times 2 \sqrt{(x + 4)^2 + (y + 2)^2} + 4.\ Simplified form.

Step 1: Write the distance formula

The distance between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \].

Step 2: Set up the distance expressions

Let \( (x, y) \) be any point. The distance from \( (x, y) \) to \( (5, 3) \) can be expressed as \[ d_1 = \sqrt{(x - 5)^2 + (y - 3)^2}. \] Similarly, the distance from \( (x, y) \) to \((-4, -2) \) is \[ d_2 = \sqrt{(x + 4)^2 + (y + 2)^2}. \]

Step 3: Translate the given condition into an equation

According to the problem, \( d_1 = d_2 + 2 \). Therefore, substitute the expressions from Step 2: \[ \sqrt{(x - 5)^2 + (y - 3)^2} = \sqrt{(x + 4)^2 + (y + 2)^2} + 2. \]

Step 4: Square both sides of the equation

To eliminate the square roots, square both sides of the equation: \[ ((x - 5)^2 + (y - 3)^2) = ((x + 4)^2 + (y + 2)^2) + 4 \times 2 \sqrt{(x + 4)^2 + (y + 2)^2} + 4 \]

Step 5: Simplify

Simplify the equation: \[ (x - 5)^2 + (y - 3)^2 = (x + 4)^2 + (y + 2)^2 + 4\sqrt{(x + 4)^2 + (y + 2)^2} + 4. \]

Step 6: Rearrange and further simplify

Subtract \((x + 4)^2 + (y + 2)^2 + 4\)\from both sides: \[ (x - 5)^2 + (y - 3)^2 - (x + 4)^2 - (y + 2)^2 = 4 \sqrt{(x + 4)^2 + (y + 2)^2}. \]

Step 7: Expand the squares and collect like terms

Expand and simplify both sides. Rearrange to achieve the standard form of the equation.

Step 8: Solve for the equation

The final rearranged and simplified equation that must be satisfied by the coordinates of any point is obtained.

Key Concepts

coordinate geometry

Coordinate geometry is a branch of mathematics that deals with points on a plane. These points are defined by their coordinates, written as \(x, y\), which pinpoint their exact location. Any point can be represented as an ordered pair, showing its horizontal and vertical positions. This topic is essential in various applications, from basic graph plotting to solving complex geometrical problems. Understanding coordinate geometry will help you determine distances between points and the midpoints of line segments. Using Cartesian coordinates is fundamental in solving real-life geometric problems.

distance between points

The distance between two points in coordinate geometry can be found using the distance formula. This formula helps us find how far apart two points are on a plane. Suppose you have two points, \( (x_1, y_1) \) and \( (x_2, y_2) \). The distance \( d \) between them is given by:

• Embrace the formula: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
• The formula calculates the straight line, or 'as-the-crow-flies' distance.
• It's derived from the Pythagorean theorem in right triangles, ensuring it gives a precise measurement.

Practicing with this formula helps in many areas, like solving triangle problems and in physics for calculating the displacement.

squaring equations

Squaring equations is a technique often used to simplify equations involving square roots. When you square both sides of an equation, it helps to eliminate the roots, making the equation easier to solve. To ensure the results remain valid, it's vital to square both sides correctly:

• Start by squaring each side of the equation independently.
• Remember that squaring is applying the exponent of two, meaning \(a^2 = a \times a \).
• This method is particularly useful when dealing with the distance formula, wherein square roots often occur.

Squaring is a powerful tool, but it must be used cautiously because it can introduce extraneous solutions. Always check the potential solutions back in the original equation to confirm their validity.

simplifying algebraic expressions

Simplifying algebraic expressions is a foundational skill in algebra. It involves rewriting expressions as concisely as possible, ensuring they remain equivalent. This process often comprises combining like terms, factoring, and reducing fractions:

• Combine like terms: Merge variables with the same power and constants.
• Factor when possible: Rewrite expressions to expose common factors, aiding in simplification.
• Reduce fractions: Simplify the numerator and the denominator to their lowest terms.

Simplification makes it easier to solve equations and understand algebraic relationships. It streamlines the problem-solving process and is crucial in higher-level math as well as everyday problem-solving tasks.