The locus of the vertices of the family of parabolas y = a³x²/3 + a²x/2 - 2a is (a) xy = 105/64 (b)

Q: Wait, why can't we just differentiate the equation with respect to $x$ to find the vertex?

Differentiating $y$ with respect to $x$ gives you the slope of the tangent, $\frac{dy}{dx}$. Setting this to zero gives you the $x$-coordinate of the vertex, but it doesn't give you the $y$-coordinate directly without substituting back. Completing the square is more robust for finding the vertex form directly.

Q: Why do we need to eliminate $a$?

The locus is defined as the path traced by the vertex as $a$ varies. If $a$ remains in your final equation, you haven't found the path; you've only found the vertex of a specific parabola. Eliminating $a$ gives you the equation of the curve that all these vertices lie on.

Q: Is the locus always a hyperbola?

In this specific case, yes! When you eliminate the parameter $a$ and end up with a relationship like $xy = C$, you have arrived at the equation of a rectangular hyperbola. It's a beautiful result!

Question

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

Options:

(A) $xy = \frac{105}{64}$

(B) $xy = \frac{3}{4}$

(C) $xy = \frac{35}{16}$

(D) $xy = \frac{64}{105}$

Sigma Percentile · Accepted Answer

### The Dance of the Parabolas

Imagine standing in a vast, empty field, watching a family of parabolas bloom into existence. Each one is unique, defined by a single, shifting parameter $a$. As $a$ changes, the parabola stretches, shifts, and morphs.

But there is a hidden order to this chaos. Every single one of these parabolas has a vertex, a 'peak' or 'valley' that marks its turning point. If you were to trace the path of these vertices as $a$ varies, what shape would you see? That is the question we are solving today.

### Phase 1: The Vertex Hunt

To find the vertex $(h, k)$ of any parabola, we need to look at it through the lens of the vertex form: $y - k = A(x - h)^2$. Our given equation is $y = \frac{a^3x^2}{3} + \frac{a^2x}{2} - 2a$.

It looks a bit messy, but we are going to use the classic, powerful technique of **completing the square**. First, we group the $x$ terms and factor out the leading coefficient, $\frac{a^3}{3}$.

This gives us:

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

Now, we look at the coefficient of $x$, which is $\frac{3}{2a}$. To complete the square, we take half of this, which is $\frac{3}{4a}$, and square it to get $\frac{9}{16a^2}$. We add and subtract this inside the bracket to keep the equation balanced:

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

### Phase 2: The Parametric Dance

With the perfect square formed, we have:

$$y = \frac{a^3}{3} \left(x + \frac{3}{4a}\right)^2 - \frac{3a}{16} - 2a$$

Simplifying the constant terms, we get the final vertex form:

$$y = \frac{a^3}{3} \left(x + \frac{3}{4a}\right)^2 - \frac{35a}{16}$$

Comparing this to $y - k = A(x - h)^2$, we can immediately extract the coordinates of the vertex $(h, k)$:

$$h = -\frac{3}{4a}, \quad k = -\frac{35a}{16}$$

### The Final Revelation

Now, we reach the climax of our journey. We have the coordinates of the vertex in terms of $a$, but we want the locus—a relationship between $x$ and $y$ that is independent of $a$. We need to eliminate the parameter $a$.

From our vertex coordinates, we have $a = -\frac{3}{4x}$ and $a = -\frac{16y}{35}$. Since both expressions equal $a$, we set them equal to each other:

$$-\frac{3}{4x} = -\frac{16y}{35}$$

Canceling the negative signs and cross-multiplying, we arrive at the elegant result:

$$105 = 64xy \implies xy = \frac{105}{64}$$

This is the equation of a **rectangular hyperbola**. We started with a family of parabolas and discovered that their vertices trace out a beautiful, smooth hyperbola. Mathematics has a way of revealing hidden connections like this, and I hope you enjoyed this journey as much as I did!

Animated Solution for Mathematics - Conic Sections: The locus of the vertices of the family of parabolas $y = \frac{a ^{3} x ^{2}}{3} + \frac{a ^{2} x}{2} - 2 a$ is

Select Answer:

Visualized Solution

The Family of Parabolas

Strategy: Completing the Square

Factoring the Leading Coefficient

Preparing the Perfect Square

Forming the Square

Distributing the Coefficient

Simplifying the Constant

Extracting the Vertex

Eliminating the Parameter $a$

Expressing $a$ in terms of $y$

Equating the Expressions

The Final Locus Equation

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

The Dance of the Parabolas

Phase 1

Phase 2

The Final Revelation

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Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The locus of the vertices of the family of parabolas y=3a3x2​+2a2x​−2a is