Parabolas `(y-alpha)^(2)=4a(x-beta)and(y-alpha)^(2)=4a'(x-beta')` will have a common normal (other than the normal passing through vertex of parabola) if

Options

A. `(2(a-a'))/(beta'-beta)gt1`
B. `(2(a-a'))/(beta-beta')gt1`
C. `(2(a'-a))/(beta'+beta)gt1`
D. `(2(a-a'))/(beta'+beta)gt1`

Answer

To solve the problem, we need to determine the conditions under which the two parabolas given by the equations \((y - \alpha)^2 = 4a(x - \beta)\) and \((y - \alpha)^2 = 4a'(x - \beta')\) have a common normal (other than the normal passing through the vertex).

### Step-by-Step Solution:

1. **Write the equations of the parabolas:**
The equations of the parabolas are:
\[ (y - \alpha)^2 = 4a(x - \beta) \quad \text{(1)} \]
\[ (y - \alpha)^2 = 4a'(x - \beta') \quad \text{(2)} \]

2. **Find the equation of the normal for the first parabola:**
The normal to the first parabola can be expressed as:
\[ y - \alpha = m(x - \beta) - 2am(1 + m^2) \]
Rearranging gives:
\[ y = mx - m\beta + \alpha - 2am(1 + m^2) \quad \text{(3)} \]

3. **Find the equation of the normal for the second parabola:**
Similarly, the normal to the second parabola is given by:
\[ y - \alpha = m(x - \beta') - 2a'm(1 + m^2) \]
Rearranging gives:
\[ y = mx - m\beta' + \alpha - 2a'm(1 + m^2) \quad \text{(4)} \]

4. **Set the equations of the normals equal to each other:**
Since the two normals must represent the same line, we set equations (3) and (4) equal to each other:
\[ -m\beta + \alpha - 2am(1 + m^2) = -m\beta' + \alpha - 2a'm(1 + m^2) \]

5. **Simplify the equation:**
Canceling \(\alpha\) from both sides and rearranging gives:
\[ -m\beta + 2am(1 + m^2) = -m\beta' + 2a'm(1 + m^2) \]
Rearranging further leads to:
\[ m(\beta' - \beta) = 2(m^2)(a' - a) \]

6. **Isolate terms:**
Dividing both sides by \(m\) (assuming \(m \neq 0\)):
\[ \beta' - \beta = 2m(a' - a) \]

7. **Condition for common normal:**
For the two parabolas to have a common normal, we need:
\[ \frac{a' - a}{\beta' - \beta} < \frac{1}{2} \]
This means:
\[ a' - a \text{ and } \beta' - \beta \text{ must have the same sign.} \]

### Conclusion:
The condition for the two parabolas to have a common normal (other than the normal passing through the vertex) is:
\[ \frac{a' - a}{\beta' - \beta} < \frac{1}{2} \]