Key Concepts and Formulas

This problem is a beautiful blend of coordinate geometry, quadratic equations, and sequences and series. To solve it effectively, we will utilize the following core concepts:

1. Intersection of Curves: If two curves intersect at a point $(x_0, y_0)$, then this point must satisfy the equations of both curves simultaneously.
2. Quadratic Equations and Discriminant: For a quadratic equation of the form $Ax^2 + Bx + C = 0$:
   • The nature of its roots is determined by the discriminant, $\Delta = B^2 - 4AC$.
   • If $\Delta = 0$, the quadratic has exactly one real root (a repeated root). This implies that the quadratic expression is a perfect square trinomial, i.e., it can be factored as $A(x-r)^2 = 0$. Geometrically, if this quadratic arises from the intersection of a line and a parabola, $\Delta = 0$ signifies that the line is tangent to the parabola.
3. Geometric Progression (G.P.): Three non-zero numbers $P, Q, R$ are in G.P. if the ratio of consecutive terms is constant, which means $\frac{Q}{P} = \frac{R}{Q}$, or equivalently, $Q^2 = PR$. If $P, Q, R$ are positive, then $Q = \sqrt{PR}$.
4. Arithmetic Progression (A.P.): Three numbers $P, Q, R$ are in A.P. if the difference between consecutive terms is constant, meaning $Q - P = R - Q$, or equivalently, $2Q = P + R$.

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Step-by-Step Solution

1. Setting Up Equations from the Intersection Condition

We are given two parabolas: $P_1: a x^2+2 b x+c y=0$ $P_2: d x^2+2 e x+f y=0$ The problem states that these parabolas intersect on the line $y=1$.

• Explanation: The phrase "intersect on the line $y=1$" means that any point of intersection $(x, y)$ must have its $y$-coordinate equal to $1$. Let's denote the $x$-coordinate of such an intersection point as $\alpha$. So, the intersection point is $(\alpha, 1)$. For this point to lie on both parabolas, its coordinates must satisfy the equations of both parabolas.

Substitute $y=1$ into the equation for Parabola $P_1$: $a \alpha^2 + 2 b \alpha + c(1) = 0$ $a \alpha^2 + 2 b \alpha + c = 0 \quad \text{(Equation 1)}$ This is a quadratic equation in $\alpha$.

Similarly, substitute $y=1$ into the equation for Parabola $P_2$: $d \alpha^2 + 2 e \alpha + f(1) = 0$ $d \alpha^2 + 2 e \alpha + f = 0 \quad \text{(Equation 2)}$ This is another quadratic equation in $\alpha$.

2. Leveraging the G.P. Condition for the First Parabola

We are given that $a, b, c$ are positive real numbers and are in Geometric Progression (G.P.).

• Explanation: By the definition of a G.P., if $a, b, c$ are in G.P., then the square of the middle term equals the product of the first and third terms. Since $a, b, c$ are positive, we can take the positive square root for $b$. Thus, we have: $b^2 = ac$ $b = \sqrt{ac}$

Now, let's substitute this relationship into Equation 1: $a \alpha^2 + 2 (\sqrt{ac}) \alpha + c = 0$

• Explanation: We observe that this quadratic equation has a special structure. Let's rewrite the terms to highlight it: $(\sqrt{a}\alpha)^2 + 2 (\sqrt{a}\alpha)(\sqrt{c}) + (\sqrt{c})^2 = 0$ This is a perfect square trinomial, which can be factored as: $(\sqrt{a} \alpha + \sqrt{c})^2 = 0$
• Alternative Explanation (using Discriminant): For Equation 1, $A=a$, $B=2b$, $C=c$. The discriminant is $\Delta = B^2 - 4AC = (2b)^2 - 4ac = 4b^2 - 4ac$. Since $b^2=ac$, we have $\Delta = 4ac - 4ac = 0$. A discriminant of zero means the quadratic has exactly one real root (a repeated root), which implies the expression is a perfect square. Geometrically, this means the line $y=1$ is tangent to the first parabola $ax^2+2bx+cy=0$ at the point $(\alpha, 1)$.

Solving for $\alpha$ from the perfect square equation: $\sqrt{a} \alpha + \sqrt{c} = 0$ $\sqrt{a} \alpha = - \sqrt{c}$ $\alpha = - \frac{\sqrt{c}}{\sqrt{a}} = - \sqrt{\frac{c}{a}}$ So, the $x$-coordinate of the unique intersection point (and tangency point) is $\alpha = - \sqrt{\frac{c}{a}}$.

3. Applying the Unique Intersection Point to the Second Parabola

The point $(\alpha, 1)$ is an intersection point for both parabolas. Therefore, the specific value of $\alpha$ we found from the properties of the first parabola must also satisfy Equation 2, which relates to the second parabola.

• Explanation: This step is crucial. By substituting the determined value of $\alpha$ into the second parabola's equation, we establish a direct link between the coefficients of the two parabolas.

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

• Careful with the square: $(-\sqrt{X})^2 = X$. So, $\left( - \sqrt{\frac{c}{a}} \right)^2 = \frac{c}{a}$. $d \left( \frac{c}{a} \right) - 2 e \sqrt{\frac{c}{a}} + f = 0$

4. Algebraic Manipulation to Discover the A.P. Relationship

Our goal is to determine if the ratios $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P. or G.P. We need to manipulate the equation obtained in Step 3 to match the form of an A.P. ($P+R=2Q$) or G.P. ($Q^2=PR$).

Let's rearrange the equation from Step 3: $\frac{dc}{a} + f = 2 e \sqrt{\frac{c}{a}}$

• Explanation: To obtain the terms $\frac{d}{a}$ and $\frac{f}{c}$, we should divide the entire equation by $c$. Since $c$ is a positive real number (given in the problem statement), it is non-zero, so this division is valid. Divide both sides by $c$: $\frac{1}{c} \left( \frac{dc}{a} + f \right) = \frac{1}{c} \left( 2 e \sqrt{\frac{c}{a}} \right)$ $\frac{d}{a} + \frac{f}{c} = 2 e \frac{1}{c} \sqrt{\frac{c}{a}}$

Now, let's simplify the right-hand side (RHS) to introduce $b$. $\text{RHS} = 2 e \frac{1}{c} \frac{\sqrt{c}}{\sqrt{a}}$

• Explanation: We can rewrite the $c$ in the denominator as $\sqrt{c} \cdot \sqrt{c}$ to facilitate cancellation and simplify the expression. $\text{RHS} = 2 e \frac{1}{\sqrt{c}\sqrt{c}} \frac{\sqrt{c}}{\sqrt{a}} = 2 e \frac{1}{\sqrt{c}\sqrt{a}} = 2 e \frac{1}{\sqrt{ac}}$

Recall from Step 2 that $b = \sqrt{ac}$ (since $a,b,c$ are positive). Substitute this into the simplified RHS: $\text{RHS} = 2 e \frac{1}{b} = 2 \frac{e}{b}$

Now, substitute this back into our main equation: $\frac{d}{a} + \frac{f}{c} = 2 \frac{e}{b}$

5. Concluding the A.P. Relationship

The final equation we derived is: $\frac{d}{a} + \frac{f}{c} = 2 \frac{e}{b}$

• Explanation: This equation perfectly matches the definition of an Arithmetic Progression. If three numbers $P, Q, R$ are in A.P., then the sum of the first and third terms is twice the middle term ($P+R = 2Q$). In our derived equation, we can identify: $P = \frac{d}{a}$ $Q = \frac{e}{b}$ $R = \frac{f}{c}$

Therefore, the terms $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

The correct option is (A).

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Tips and Common Mistakes

• **Understanding "Intersect on the