Problem 42 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 41-42: Find an equation for the parabola that has a horizontal axis and passes through the given points. $$ P(2,1), \quad Q(6,2), \quad R(12,-1) $$

Step-by-Step Solution

Answer

The equation of the parabola is \(x = 3y^2 - 5y + 4\).

Step 1: Understanding Parabola Equation

To find the equation of a parabola with a horizontal axis, use the general form: \(x = ay^2 + by + c\). This equation indicates the parabola opens either to the right or to the left.

Step 2: Substitute Points into Equation

Substitute the given points into the equation \(x = ay^2 + by + c\). Each point will provide an equation: \((2 = a(1)^2 + b(1) + c)\), \((6 = a(2)^2 + b(2) + c)\), and \((12 = a(-1)^2 + b(-1) + c)\).

Step 3: Simplify and Form Equations

After simplification, the equations from the points become: \(2 = a + b + c\), \(6 = 4a + 2b + c\), and \(12 = a - b + c\). These are a system of equations that need solving.

Step 4: Solve the System of Equations

Solve the system of equations: 1. Subtract the first equation from the second: \(6 - 2 = 4a + 2b + c - (a + b + c)\) resulting in \(4 = 3a + b\).2. Subtract the first equation from the third: \(12 - 2 = a - b + c - (a + b + c)\) gives \(10 = -2b\) leading to \(b = -5\).3. Substitute \(b = -5\) into \(4 = 3a + b\) results in \(4 = 3a - 5\), so \(3a = 9\) resulting in \(a = 3\).4. Substitute \(a = 3\) and \(b = -5\) into \(2 = a + b + c\), then \(2 = 3 - 5 + c\) so \(c = 4\).

Step 5: Substitute Parameters into Equation

Substitute \(a = 3\), \(b = -5\), and \(c = 4\) back into the general equation \(x = ay^2 + by + c\). The equation of the parabola is \(x = 3y^2 - 5y + 4\).

Key Concepts

System of Equations

A system of equations involves multiple equations with multiple variables. We solve them simultaneously for one unique solution set. In the context of finding the equation of a parabola, we derive a system of equations by plugging the points into the general form of the parabola equation, which is determined by the relationship of the points to the axis of the parabola.
For the parabola with a horizontal axis, the general form is \(x = ay^2 + by + c\).
This form creates different equations when specific points known to lie on the parabola are substituted.Here is how a system is formed:

• Take each point provided (like P(2,1), Q(6,2), R(12,-1)) and substitute it into the general equation \(x = ay^2 + by + c\).
• Each substitution results in a unique equation: hence for P(2,1), you get \(2 = a(1)^2 + b(1) + c\).
• Repeat the process for all given points, forming a complete system of simultaneous equations.

The simultaneous solutions give the values for \(a\), \(b\), and \(c\). These constants define the specific parabola fitting through all the given points.

Horizontal Axis

A parabola with a horizontal axis means that it opens either to the left or to the right. This characteristic requires the general equation to be in the form of \(x = ay^2 + by + c\), differing from the vertical oepening, which would be represented by \(y = ax^2 + bx + c\).
Key aspects of a horizontal axis:

• The \(y\)-terms in the equation multiply and sum up to affect the \(x\) values. Thus, their resolution in the system of equations manipulates how the graph opens sideways.
• The presence of the \(y^2\) term assures one that the structure is parabolic, and not simply linear or any other form.

Understanding this directional opening is essential when determining the exact nature of the parabola with help from point substitutions.

Solving Algebraic Equations

When faced with a system of algebraic equations, solving involves finding values for unknown variables that satisfy all equations simultaneously. This is a methodology frequently seen in geometry and algebra alike. In our problem, where we substitute the specific points into a generalized parabola equation, we derive three primary equations. Steps to solve the equations:

• Rearrange or transform the equations to isolate and simplify terms, often done by subtracting similar equations from each other as shown in our exercise step-by-step.
• Methodically solve for one variable at a time. For example, solve for \(b\) in one equation and substitute back into another to isolate \(a\).
• Once individual constants are identified, substitute them back into the simplest of the equations to validate and solve for the last unknown, \(c\).

Algorithmically solving these is fundamental, often using elimination or substitution methods, and ultimately finding the specific expression defining the parabola.

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebraic important concepts with geometric figures. It utilizes a coordinate system—you typically employ Cartesian coordinates—to refer all points uniquely. This allows for a detailed algebraic understanding of shape properties, positioned in a plane. In the question given, we are using coordinate geometry to understand spatial orientations:

• Every point like P(2,1) is represented with exact numerical references to its place on the plane, facilitating its use in equations.
• By representing such points algebraically, we convert questions of shape and position into solvable algebraic expressions.

The essential beauty of coordinate geometry lies in creating tangible mathematical representations of spatial relationships, which enhances problem-solving for geometric figures such as parabolas.