The parabolas $y^2 = 4ax$ and $y^2 = 4c(x - d)$ have a common normal other than the X-axis if and only if:
$c > a$ and $2a > d + 2c$
$c d + 2c$
$c > a$ and $2a < d + 2c$
$c < a$ and $2a < d + 2c$
Step-by-Step Solution
Key Concept: For two parabolas y² = 4ax and y² = 4c(x - d) to share a common normal (other than x-axis), the normal line y = mx - 2am - am³ from the first parabola must coincide with y = m(x - d) - 2cm - cm³ from the second, requiring -2am - am³ = -m(d) - 2cm - cm³, which simplifies to m²(a - c) = 2(c - a)d, yielding conditions on the relationship between a, c, and d.
The normal to $y^2 = 4ax$ at point $(at_0^2, 2at_0)$ has equation $y = mx - 2am - am^3$ where $m$ is the slope. For two parabolas $y^2 = 4ax$ and $y^2 = 4c(x - d)$, a normal with the same slope $m$ to the second parabola is $y = m(x - d) - 2cm - cm^3$. These represent the same line when $-2am - am^3 = -dm - 2cm - cm^3$, giving either $m = 0$ (x-axis) or $m^2 = \frac{2a - d - 2c}{c - a}$. For real non-zero slopes, $\frac{2a - d - 2c}{c - a} > 0$.
Correct Answer: 1,4