Three normals drawn to the parabola `y^(2)` = 4x from the point (c, 0) are real and diferent if

Options

A. c=0
B. c=1
C. c=2
D. c=3

Answer

To determine the conditions under which three normals can be drawn to the parabola \( y^2 = 4x \) from the point \( (c, 0) \), we can follow these steps:

### Step 1: Identify the standard form of the parabola
The given parabola is \( y^2 = 4x \). This can be compared with the standard form \( y^2 = 4ax \), where \( a = 1 \).

### Step 2: Write the equation of the normal
The equation of the normal to the parabola at a point where the slope is \( m \) is given by:
\[ y = mx - 2am + a^3 \]
Substituting \( a = 1 \):
\[ y = mx - 2m + 1 \]

### Step 3: Substitute the point \( (c, 0) \) into the normal equation
Since the point \( (c, 0) \) lies on the normal, we substitute \( x = c \) and \( y = 0 \):
\[ 0 = mc - 2m + 1 \]
Rearranging gives:
\[ mc - 2m + 1 = 0 \]

### Step 4: Factor out \( m \)
Factoring out \( m \):
\[ m(c - 2) + 1 = 0 \]
This can be rearranged to:
\[ m(c - 2) = -1 \]

### Step 5: Solve for \( m \)
From the above equation, we can express \( m \):
\[ m = \frac{-1}{c - 2} \]

### Step 6: Determine conditions for real and distinct normals
For three distinct normals, we need to ensure that the quadratic equation in \( m \) has three distinct solutions. The quadratic equation derived from the normal condition is:
\[ m^2 + 2m - c = 0 \]
This is a quadratic equation in \( m \). For it to have real and distinct roots, the discriminant must be greater than zero:
\[ D = b^2 - 4ac = 2^2 - 4(1)(-c) > 0 \]
This simplifies to:
\[ 4 + 4c > 0 \]
Thus:
\[ c > -1 \]

### Step 7: Analyze the condition for three distinct normals
From our earlier step, we also need to ensure that \( c - 2 \neq 0 \) (to avoid a double root), which means:
\[ c \neq 2 \]
To have three distinct normals, we require:
1. \( c > 2 \) (to ensure \( m^2 - c \) has two real roots).
2. \( c \neq 2 \).

### Conclusion
Thus, the condition for three normals to be real and different is:
\[ c > 2 \]