If the line y=mx+c touches the parabola `y^(2)=4a(x+a)`, then

Options

A. `c=a+a/m`
B. `c=am+a/m`
C. `c=am+a`
D. none of these

Explanation

To find the condition under which the line \( y = mx + c \) touches the parabola \( y^2 = 4a(x + a) \), we can follow these steps:

### Step 1: Substitute the line equation into the parabola equation
We start by substituting \( y = mx + c \) into the parabola equation \( y^2 = 4a(x + a) \).

\[ (mx + c)^2 = 4a(x + a) \]

### Step 2: Expand both sides
Expanding the left-hand side, we have:

\[ m^2x^2 + 2mcx + c^2 = 4ax + 4a^2 \]

### Step 3: Rearrange the equation
Rearranging gives us:

\[ m^2x^2 + (2mc - 4a)x + (c^2 - 4a^2) = 0 \]

### Step 4: Use the condition for tangency
For the line to be tangent to the parabola, the quadratic equation must have exactly one solution. This occurs when the discriminant is zero. The discriminant \( D \) of the quadratic equation \( Ax^2 + Bx + C = 0 \) is given by:

\[ D = B^2 - 4AC \]

In our case, \( A = m^2 \), \( B = 2mc - 4a \), and \( C = c^2 - 4a^2 \). Therefore, we set the discriminant to zero:

\[ (2mc - 4a)^2 - 4(m^2)(c^2 - 4a^2) = 0 \]

### Step 5: Simplify the discriminant equation
Expanding the discriminant, we have:

\[ (2mc - 4a)^2 = 4m^2(c^2 - 4a^2) \]

### Step 6: Expand both sides
Expanding the left side gives:

\[ 4m^2c^2 - 16mac + 16a^2 = 4m^2c^2 - 16a^2m^2 \]

### Step 7: Cancel common terms
Now, we can cancel \( 4m^2c^2 \) from both sides:

\[ -16mac + 16a^2 = -16a^2m^2 \]

### Step 8: Rearranging the equation
Rearranging gives us:

\[ 16a^2m^2 - 16mac + 16a^2 = 0 \]

### Step 9: Factor out common terms
Factoring out \( 16a \):

\[ 16a(am - mc + a) = 0 \]

### Step 10: Solve for \( c \)
Since \( a \neq 0 \), we can divide by \( 16a \):

\[ am - mc + a = 0 \]

Rearranging gives:

\[ mc = a + am \]

Dividing by \( m \) (assuming \( m \neq 0 \)):

\[ c = \frac{a}{m} + a m \]

### Final Condition
Thus, the condition for the line \( y = mx + c \) to touch the parabola \( y^2 = 4a(x + a) \) is:

\[ c = \frac{a}{m} + am \]