The parabolas : ax² + 2bx + cy = 0 and dx² + 2ex + fy = 0 intersect on the line y = 1. If a, b, c, d

Q: Wait, why do we assume the parabolas intersect at only one point? Could there be two?

That is a brilliant observation! By forcing the first equation to be a perfect square, we found that it has a unique root. This means the parabola $ax^2 + 2bx + cy = 0$ is tangent to the line $y=1$ at that specific $x$. Since the second parabola also passes through this point, they are essentially 'kissing' at that coordinate. We don't need to worry about other intersection points because the problem implies a shared condition at $y=1$.

Q: Why did we divide by $c$ in the final steps? Is that always allowed?

Great question! The problem explicitly states that $a, b, c, d, e, f$ are positive real numbers. Since $c > 0$, it is perfectly safe to divide by $c$ without worrying about division by zero or flipping inequality signs. It is a strategic move to normalize the equation into a form that reveals the A.P. structure.

Q: I got stuck at $\frac{dc}{a} - \frac{2e\sqrt{c}}{\sqrt{a}} + f = 0$. How do I turn that into the A.P. condition?

You were so close! The trick is to look at the target expression $\frac{d}{a} + \frac{f}{c} = \frac{2e}{b}$. Notice that your term $\frac{dc}{a}$ needs to become $\frac{d}{a}$. Dividing the entire equation by $c$ is the key that unlocks this. Once you divide, the middle term $\frac{2e\sqrt{c}}{c\sqrt{a}}$ simplifies beautifully to $\frac{2e}{\sqrt{ac}}$, and since $\sqrt{ac} = b$, you arrive exactly at the A.P. definition.

Q: Can I use the discriminant method instead of perfect squares?

You absolutely could! If the parabolas intersect at $y=1$, the equation $ax^2 + 2bx + c = 0$ must have a discriminant $D = (2b)^2 - 4ac = 0$. Since $b^2 = ac$, this confirms $D=0$. While this confirms the root is unique, the perfect square method is faster and more elegant for algebraic manipulation.

Q: Is there a geometric interpretation for why they are in A.P.?

Think of the A.P. condition as a 'balance' of ratios. The ratios $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ represent how the coefficients of the second parabola scale relative to the first. The fact that they form an A.P. suggests a linear relationship between these scaling factors, which is a beautiful consequence of the parabolas sharing a common point of tangency on the line $y=1$.

Question

The parabolas : $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on the line $y = 1$. If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in G.P., then

Options:

(A) $d, e, f$ are in A.P.

(B) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in G.P.

(C) $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

(D) $d, e, f$ are in G.P.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before two complex, shifting curves—two parabolas defined by the equations $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$. At first glance, they seem like independent entities wandering through the Cartesian plane.

However, there is a secret tether binding them: they are destined to meet on the horizontal line $y = 1$. Today, we are going to uncover the hidden harmony between their coefficients.

### The Point of Contact

When we say two curves intersect at a specific line, we are essentially saying that at that exact moment, they share a common coordinate. If they meet at $y = 1$, then for both equations, the variable $y$ is no longer a variable—it is a constant.

By substituting $y = 1$ into our equations, we collapse the two-dimensional problem into a one-dimensional search for the intersection point $x$:

$$ax^2 + 2bx + c = 0$$

$$dx^2 + 2ex + f = 0$$

These are the 'snapshots' of our parabolas at the moment of intersection. We are told that $a, b, c$ are in a Geometric Progression (G.P.), which serves as a mathematical invitation to simplify the first equation.

### The Perfect Square Revelation

In a G.P., the middle term is the geometric mean of its neighbors, meaning $b^2 = ac$, or $b = \sqrt{ac}$. Let us inject this truth into our first equation:

$$ax^2 + 2\sqrt{ac}x + c = 0$$

If we let $p = \sqrt{a}x$ and $q = \sqrt{c}$, the equation becomes $p^2 + 2pq + q^2 = 0$. This is the classic expansion of $(p + q)^2$. Thus, our equation simplifies to:

$$(\sqrt{a}x + \sqrt{c})^2 = 0$$

This tells us that the intersection point is a point of tangency where $x = -\sqrt{\frac{c}{a}}$. The parabolas are not just crossing; they are touching the line $y=1$ with perfect precision.

### The Bridge to Arithmetic Progression

Now, we take this value of $x$ and carry it over to the second parabola. We substitute $x = -\sqrt{\frac{c}{a}}$ into $dx^2 + 2ex + f = 0$:

$$d\left(-\sqrt{\frac{c}{a}}\right)^2 + 2e\left(-\sqrt{\frac{c}{a}}\right) + f = 0$$

Simplifying this, we obtain:

$$\frac{dc}{a} - \frac{2e\sqrt{c}}{\sqrt{a}} + f = 0$$

To simplify further, we divide the entire equation by $c$:

$$\frac{d}{a} - \frac{2e\sqrt{c}}{c\sqrt{a}} + \frac{f}{c} = 0$$

Since $\frac{\sqrt{c}}{c} = \frac{1}{\sqrt{c}}$, the middle term becomes $\frac{2e}{\sqrt{ac}}$. Because we know $\sqrt{ac} = b$, we can rewrite the entire expression as:

$$\frac{d}{a} - \frac{2e}{b} + \frac{f}{c} = 0$$

### The Grand Finale

Rearranging the terms, we arrive at the final, elegant result:

$$\frac{d}{a} + \frac{f}{c} = 2\left(\frac{e}{b}\right)$$

This is the hallmark of an Arithmetic Progression. If the sum of the first and third terms is twice the middle term, then the terms themselves must be in A.P.

We have successfully proven that **$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.** This result demonstrates the hidden architecture linking the coefficients of these two parabolas.

Animated Solution for Mathematics - Quadratic Equations: The parabolas : $a x^{2} + 2 b x + cy = 0$ and $d x^{2} + 2 e x + f y = 0$ intersect on the line $y = 1$ . If $a, b, c, d, e, f$ are positive real numbers and $a, b, c$ are in G.P., then

Select Answer:

Visualized Solution

Visualizing the Intersection

Substituting $y = 1$

Applying G.P. Condition

Forming a Perfect Square

Solving for $x$

Substituting $x$ in Second Equation

Simplifying the Expression

Normalizing the Equation

Final Substitution using $b$

Identifying the Progression

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Point of Contact

The Perfect Square Revelation

The Bridge to Arithmetic Progression

The Grand Finale

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