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Step-by-Step Solution
Step 1: Set up the equations of the parabolas at y = 1
Each parabola is given by:
First parabola: $a x^2 + 2 b x + c y = 0$
Second parabola: $d x^2 + 2 e x + f y = 0$
We are told they intersect at $y = 1$. Substituting $y = 1$ into each equation gives:
$a \alpha^2 + 2 b \alpha + c = 0 \quad \dots (1)$
$d \alpha^2 + 2 e \alpha + f = 0 \quad \dots (2)$
Step 2: Use the information that $a, b, c$ are in geometric progression (G.P.)
Since $a, b, c$ are in G.P., it means $b^2 = a \cdot c$. Hence,
$b = \sqrt{a c}.
Rewriting equation (1) with $b = \sqrt{ac}$:
$a \alpha^2 + 2 \sqrt{ac} \,\alpha + c = 0.
Step 3: Solve for $x = \alpha$ from the first parabola
Group the terms in a convenient way:
$a \alpha^2 + 2 \sqrt{ac} \,\alpha + c = (\sqrt{a}\,\alpha + \sqrt{c})^2 = 0.
Since a perfect square equals zero, we get:
$\sqrt{a}\,\alpha + \sqrt{c} = 0 \quad \Longrightarrow \quad \alpha = -\,\sqrt{\frac{c}{a}}.
Step 4: Substitute $\alpha$ into the second parabola’s equation
Now, plug $\alpha = -\sqrt{\tfrac{c}{a}}$ into
$d\alpha^2 + 2 e \alpha + f = 0.
First, note that $\alpha^2 = \frac{c}{a}$. So,
$d \left(\frac{c}{a}\right) + 2 e \left(-\,\sqrt{\frac{c}{a}}\right) + f = 0.
Rearranging terms,
$\frac{d}{a} \, c - 2 e \sqrt{\frac{c}{a}} + f = 0.
Divide through by $c$ (assuming $c > 0$) to simplify:
$\frac{d}{a} - \frac{2 e}{\sqrt{a c}} + \frac{f}{c} = 0.
Step 5: Recognize the arithmetic progression relation
Recall that $b = \sqrt{ac}$, so $\sqrt{ac} = b$. Therefore,
$\frac{d}{a} + \frac{f}{c} = 2\,\frac{e}{b}.
This is precisely the condition that the three numbers $\frac{d}{a}$, $\frac{e}{b}$, and $\frac{f}{c}$ form an arithmetic progression, because in an A.P. the middle term is the average of the first and the third:
$\frac{e}{b} = \frac{1}{2} \left( \frac{d}{a} + \frac{f}{c} \right).
Hence,
$\frac{d}{a}, \frac{e}{b}, \frac{f}{c}
are in an arithmetic progression.
