Understanding the Problem and Key Concepts

This problem asks us to find the locus of the vertices of a family of parabolas.

• A "family of parabolas" means we have an equation that defines many parabolas, each determined by a specific value of a parameter (in this case, '$a$').
• The "locus" of the vertices refers to the path traced by these vertices as the parameter '$a$' varies. To find this locus, we generally need to:
  1. Determine the coordinates of the vertex $(h,k)$ in terms of the parameter '$a$'.
  2. Eliminate the parameter '$a$' from these coordinate equations to obtain a single equation relating $h$ and $k$. This equation will be the locus.

Key Concept: Vertex of a Parabola

For a parabola given by the general quadratic equation $y = Ax^2 + Bx + C$ (which opens upwards if $A>0$ or downwards if $A<0$), its vertex $(h,k)$ can be found using two primary methods:

1. Direct Formula: The x-coordinate of the vertex is $h = -\frac{B}{2A}$. The corresponding y-coordinate $k$ is found by substituting this value of $h$ back into the parabola's equation: $k = A\left(-\frac{B}{2A}\right)^2 + B\left(-\frac{B}{2A}\right) + C$.
2. Completing the Square: Rewrite the equation in the standard form $(x-h)^2 = 4p(y-k)$. Once in this form, the vertex $(h,k)$ can be directly identified. This method is often preferred as it also reveals the focal length and direction of opening, providing a deeper insight into the parabola's structure.

In this solution, we will demonstrate the completing the square method.

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Step-by-Step Solution

Step 1: Identify the Equation of the Parabola and its Coefficients

The given equation of the family of parabolas is: $y = \frac{a^3}{3}x^2 + \frac{a^2}{2}x - 2a$

Why this step? This is the crucial first step. By comparing the given equation to the general form $y = Ax^2 + Bx + C$, we clearly identify the coefficients $A$, $B$, and $C$ in terms of the parameter '$a$'. This identification is essential for applying any method to find the vertex. Here, we have: $A = \frac{a^3}{3}$ $B = \frac{a^2}{2}$ $C = -2a$

Step 2: Find the Coordinates of the Vertex $(h,k)$ using Completing the Square

Our goal is to transform the equation into the standard form $(x-h)^2 = 4p(y-k)$.

First, to begin completing the square for the $x$ terms, we factor out the coefficient of $x^2$ (which is $A$) from the terms involving $x$: $y = A \left( x^2 + \frac{B}{A}x \right) + C$ Substituting our identified $A$, $B$, and $C$: $y = \frac{a^3}{3} \left( x^2 + \frac{a^2/2}{a^3/3}x \right) - 2a$ Simplify the coefficient of $x$ inside the parenthesis: $y = \frac{a^3}{3} \left( x^2 + \frac{a^2}{2} \cdot \frac{3}{a^3}x \right) - 2a$ $y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a}x \right) - 2a$

Why this step? Factoring out the leading coefficient ($A$) from the $x^2$ and $x$ terms makes the process of completing the square much simpler. We aim to create a perfect square trinomial of the form $(x+m)^2 = x^2 + 2mx + m^2$ within the parenthesis.

Next, we complete the square for the expression $x^2 + \frac{3}{2a}x$. To do this, we add and subtract the square of half the coefficient of $x$. The coefficient of $x$ is $\frac{3}{2a}$, so half of it is $\frac{1}{2} \cdot \frac{3}{2a} = \frac{3}{4a}$. The term to add and subtract is $\left(\frac{3}{4a}\right)^2 = \frac{9}{16a^2}$.

$y = \frac{a^3}{3} \left( x^2 + \frac{3}{2a}x + \frac{9}{16a^2} - \frac{9}{16a^2} \right) - 2a$

Why this step? Adding and subtracting the same value within the parenthesis ensures that the overall equation remains unchanged. This allows us to group the first three terms to form a perfect square trinomial.

Now, group the perfect square trinomial and separate the subtracted term: $y = \frac{a^3}{3} \left[ \left(x + \frac{3}{4a}\right)^2 - \frac{9}{16a^2} \right] - 2a$

Distribute the $\frac{a^3}{3}$ term back into the parenthesis: $y = \frac{a^3}{3} \left(x + \frac{3}{4a}\right)^2 - \left(\frac{a^3}{3} \cdot \frac{9}{16a^2}\right) - 2a$

Tip: A common mistake here is to forget to multiply the subtracted term ($\frac{9}{16a^2}$) by the coefficient that was factored out ($\frac{a^3}{3}$). This leads to an incorrect constant term.

Simplify the constant terms: $y = \frac{a^3}{3} \left(x + \frac{3}{4a}\right)^2 - \frac{3a}{16} - 2a$ Combine the constant terms by finding a common denominator: $y = \frac{a^3}{3} \left(x + \frac{3}{4a}\right)^2 - \left(\frac{3a}{16} + \frac{32a}{16}\right)$ $y = \frac{a^3}{3} \left(x + \frac{3}{4a}\right)^2 - \frac{35a}{16}$

Finally, rearrange the equation to match the standard form $(x-h)^2 = 4p(y-k)$: $y + \frac{35a}{16} = \frac{a^3}{3} \left(x + \frac{3}{4a}\right)^2$ To isolate the squared term, multiply both sides by $\frac{3}{a^3}$: $\left(x + \frac{3}{4a}\right)^2 = \frac{3}{a^3} \left(y + \frac{35a}{16}\right)$

Comparing this to the standard form $(x-h)^2 = 4p(y-k)$, we can identify the coordinates of the vertex $(h,k)$: $h = -\frac{3}{4a}$ $k = -\frac{35a}{16}$

Step 3: Express Vertex Coordinates in Terms of $X$ and $Y$ for Locus

Let the coordinates of the vertex be $(X, Y)$. This is a common convention to distinguish the locus coordinates from the general variables $x$ and $y$ in the parabola's equation. So, we have two parametric equations for the locus: $X = -\frac{3}{4a} \quad \text{ (Equation 1)}$ $Y = -\frac{35a}{16} \quad \text{ (Equation 2)}$

Why this step? We have now found the coordinates of any vertex of the family of parabolas in terms of the parameter '$a$'. To find the locus, we need to find a direct relationship between these $X$ and $Y$ coordinates that is independent of '$a$'.

Step 4: Eliminate the Parameter '$a$' to Find the Locus

To eliminate '$a$', we can solve one equation for '$a$' and substitute it into the other equation. From Equation 1, solve for $a$: $4aX = -3$ $a = -\frac{3}{4X}$

Now, substitute this expression for $a$ into Equation 2: $Y = -\frac{35}{16} \left(-\frac{3}{4X}\right)$

Why this step? This is the core step in finding the locus. By substituting $a$ from one equation into the other, we obtain an equation that relates $X$ and $Y$ directly, thereby defining the path (locus) traced by the vertex.

Simplify the expression for $Y$: $Y = \frac{35 \times 3}{16 \times 4X}$ $Y = \frac{105}{64X}$

Finally, multiply both sides by $X$ to clear the denominator and express the locus in a standard form: $XY = \frac{105}{64}$

This is the equation of the locus of the vertices of the family of parabolas.

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Summary and Key Takeaway

The locus of the vertices of the given family of parabolas is $xy = \frac{105}{64}$. This equation represents a hyperbola, specifically a rectangular hyperbola whose asymptotes are the coordinate axes.

The problem illustrates a general strategy for finding the locus of a point related to a parameterized family of curves:

1. Parameterize the Point: Express the coordinates of the point (whose locus is desired) in terms of the given parameter.
2. Eliminate the Parameter: Eliminate the parameter from these coordinate equations to obtain a single equation relating the $X$ and $Y$ coordinates. This resulting equation is the locus.