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

Options

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

Answer

To solve the problem where the line \( y = mx + c \) touches the parabola \( y^2 = 4a(x + a) \), we can follow these steps:

### Step 1: Rewrite the Parabola
The given parabola is \( y^2 = 4a(x + a) \). We can rewrite it in a more standard form:
\[ y^2 = 4a(x + a) \implies y^2 = 4ax + 4a^2 \]

### Step 2: Identify the Tangent Line
The equation of the tangent line is given as \( y = mx + c \). For this line to be a tangent to the parabola, it must satisfy the condition that the quadratic formed when substituting \( y \) from the line into the parabola has exactly one solution.

### Step 3: Substitute the Line into the Parabola
Substituting \( y = mx + c \) into the parabola's equation \( y^2 = 4a(x + a) \):
\[ (mx + c)^2 = 4a(x + a) \]
Expanding both sides:
\[ m^2x^2 + 2mcx + c^2 = 4ax + 4a^2 \]

### Step 4: Rearranging the Equation
Rearranging gives us:
\[ m^2x^2 + (2mc - 4a)x + (c^2 - 4a^2) = 0 \]

### Step 5: Condition for Tangency
For the line to be a tangent to the parabola, the discriminant of this quadratic equation must be zero:
\[ D = (2mc - 4a)^2 - 4m^2(c^2 - 4a^2) = 0 \]

### Step 6: Expanding the Discriminant
Expanding the discriminant:
\[ (2mc - 4a)^2 = 4m^2(c^2 - 4a^2) \]
This simplifies to:
\[ 4m^2c^2 - 16mac + 16a^2 = 4m^2c^2 - 16a^2m^2 \]

### Step 7: Simplifying the Equation
Cancelling \( 4m^2c^2 \) from both sides:
\[ -16mac + 16a^2 = -16a^2m^2 \]
Dividing through by -16:
\[ mac - a^2 = a^2m^2 \]

### Step 8: Rearranging for \( c \)
Rearranging gives:
\[ c = am + \frac{a^2m^2}{a} = am + \frac{a}{m} \]

### Final Result
Thus, we find that:
\[ c = am + \frac{a}{m} \]