The locus of the vertices of the family of parabolas `y =[a^3x^2]/3 + [a^2x]/2 -2a` is:

Options

A. `xy=(105)/(64)`
B. `xy=3/4`
C. `xy=35/16`
D. `xy=64/105`

Explanation

To find the locus of the vertices of the family of parabolas given by the equation \( y = \frac{a^3 x^2}{3} + \frac{a^2 x}{2} - 2a \), we will follow these steps:

### Step 1: Differentiate the equation
We start by differentiating the equation with respect to \( x \) to find the critical points (vertices).

\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{a^3 x^2}{3} + \frac{a^2 x}{2} - 2a\right) \]

Using the power rule for differentiation, we get:

\[ \frac{dy}{dx} = \frac{2a^3 x}{3} + \frac{a^2}{2} \]

### Step 2: Set the derivative to zero
To find the x-coordinate of the vertex, we set the derivative equal to zero:

\[ \frac{2a^3 x}{3} + \frac{a^2}{2} = 0 \]

### Step 3: Solve for \( x \)
Rearranging the equation gives us:

\[ \frac{2a^3 x}{3} = -\frac{a^2}{2} \]

Multiplying both sides by 6 to eliminate the fractions:

\[ 4a^3 x = -3a^2 \]

Now, divide both sides by \( 4a^3 \) (assuming \( a \neq 0 \)):

\[ x = -\frac{3}{4a} \]

### Step 4: Substitute \( x \) back into the original equation to find \( y \)
Now we substitute \( x = -\frac{3}{4a} \) back into the original equation to find the corresponding \( y \)-coordinate:

\[ y = \frac{a^3 \left(-\frac{3}{4a}\right)^2}{3} + \frac{a^2 \left(-\frac{3}{4a}\right)}{2} - 2a \]

Calculating each term:

1. The first term:
\[ \frac{a^3 \cdot \frac{9}{16a^2}}{3} = \frac{3a}{16} \]

2. The second term:
\[ \frac{-3a^2}{8a} = -\frac{3}{8} \]

3. The third term:
\[ -2a \]

Combining these terms gives us:

\[ y = \frac{3a}{16} - \frac{3}{8} - 2a \]

Finding a common denominator (16):

\[ y = \frac{3a}{16} - \frac{6a}{16} - \frac{32a}{16} = \frac{3a - 6a - 32a}{16} = \frac{-35a}{16} \]

### Step 5: Find the product \( xy \)
Now we have \( x = -\frac{3}{4a} \) and \( y = -\frac{35a}{16} \). We can find \( xy \):

\[ xy = \left(-\frac{3}{4a}\right) \left(-\frac{35a}{16}\right) = \frac{3 \cdot 35}{4 \cdot 16} = \frac{105}{64} \]

### Conclusion
The locus of the vertices of the family of parabolas is given by the equation \( xy = \frac{105}{64} \).

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