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 determine the condition under which the parabolas \((y - \alpha)^2 = 4a(x - \beta)\) and \((y - \alpha)^2 = 4a'(x - \beta')\) have a common normal (other than the normal passing through the vertex), we can follow these steps:

### Step-by-Step Solution

1. **Identify the Normal Equation**:
The normal to the parabola \((y - \alpha)^2 = 4a(x - \beta)\) at a point can be expressed as:
\[ y - \alpha = m(x - \beta) - 2a m - a m^3 \]
where \(m\) is the slope of the normal.

2. **Write the Normal for Both Parabolas**:
For the first parabola:
\[ y = mx + \left(\alpha - m\beta + 2am + am^3\right) \]
For the second parabola \((y - \alpha)^2 = 4a'(x - \beta')\):
\[ y - \alpha = m(x - \beta') - 2a'm - a'm^3 \]
which simplifies to:
\[ y = mx + \left(\alpha - m\beta' + 2a'm + a'm^3\right) \]

3. **Set the Two Normal Equations Equal**:
For the normals to be common, the right-hand sides of both equations must be equal:
\[ \alpha - m\beta + 2am + am^3 = \alpha - m\beta' + 2a'm + a'm^3 \]

4. **Cancel \(\alpha\) and Rearrange**:
Cancel \(\alpha\) from both sides:
\[ -m\beta + 2am + am^3 = -m\beta' + 2a'm + a'm^3 \]
Rearranging gives:
\[ m\beta' - m\beta + 2a'm - 2am + a'm^3 - am^3 = 0 \]

5. **Factor Out Common Terms**:
Group the terms:
\[ m(\beta' - \beta) + (2a' - 2a)m + (a' - a)m^3 = 0 \]

6. **Consider the Case for Non-trivial Solutions**:
For this equation to hold for non-trivial \(m\) (not equal to zero), we can factor out \(m\):
\[ m\left((\beta' - \beta) + (2a' - 2a) + (a' - a)m^2\right) = 0 \]
This implies:
\[ (\beta' - \beta) + (2a' - 2a) + (a' - a)m^2 = 0 \]

7. **Solve for \(m^2\)**:
Rearranging gives:
\[ m^2 = \frac{-(\beta' - \beta) - (2a' - 2a)}{(a' - a)} \]

8. **Condition for Real Solutions**:
For \(m^2\) to be non-negative, the right-hand side must be greater than or equal to zero:
\[ -(\beta' - \beta) - (2a' - 2a) \geq 0 \]
This leads to the condition:
\[ \beta' - \beta + 2(a' - a) \leq 0 \]

### Final Condition
Thus, the parabolas will have a common normal (other than the normal passing through the vertex) if:
\[ \frac{2a - a'}{\beta - \beta'} > 1 \]