If the line `x+my+am^(2)=0` touches the parabola `y^(2)= 4ax` then the ponit of contact is

Options

A. `(-am^(2),2am)`
B. `(-am^(2), -2am)`
C. `(am^(2),-2am)`
D. `(am^(2),2am)`

Answer

To find the point of contact of the line \( x + my + am^2 = 0 \) with the parabola \( y^2 = 4ax \), we can follow these steps:

### Step 1: Rearranging the Line Equation
The equation of the line can be rearranged to express \( x \) in terms of \( y \):
\[ x = -my - am^2 \]

### Step 2: Substitute into the Parabola Equation
Substituting the expression for \( x \) into the parabola equation \( y^2 = 4ax \):
\[ y^2 = 4a(-my - am^2) \]
This simplifies to:
\[ y^2 = -4amy - 4a^2m^2 \]

### Step 3: Rearranging to Form a Quadratic Equation
Rearranging the equation gives:
\[ y^2 + 4amy + 4a^2m^2 = 0 \]

### Step 4: Applying the Discriminant Condition
For the line to be tangent to the parabola, the discriminant of this quadratic equation must be zero. The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = 1 \), \( b = 4am \), and \( c = 4a^2m^2 \). Thus,
\[ D = (4am)^2 - 4 \cdot 1 \cdot 4a^2m^2 \]
Calculating this gives:
\[ D = 16a^2m^2 - 16a^2m^2 = 0 \]

### Step 5: Finding the Value of \( y \)
Since the discriminant is zero, there is exactly one solution for \( y \). We can factor the quadratic:
\[ (y + 2am)^2 = 0 \]
This implies:
\[ y + 2am = 0 \implies y = -2am \]

### Step 6: Finding the Corresponding \( x \)
Now substituting \( y = -2am \) back into the expression for \( x \):
\[ x = -m(-2am) - am^2 = 2am^2 - am^2 = am^2 \]

### Step 7: Point of Contact
Thus, the point of contact is:
\[ \text{Point of contact} = (am^2, -2am) \]

### Final Answer
The point of contact of the line \( x + my + am^2 = 0 \) with the parabola \( y^2 = 4ax \) is:
\[ (am^2, -2am) \]