The line y = mx + 1 is a tangent to the parabola `y^2 = 4x `

Options

A. `m=1 `
B. `m=2`
C. `m=4`
D. ` m=3 `

Answer

To determine the value of \( m \) for which the line \( y = mx + 1 \) is a tangent to the parabola \( y^2 = 4x \), we can follow these steps:

### Step 1: Write the equations
We have the parabola given by the equation:
\[ y^2 = 4x \]
And the line given by:
\[ y = mx + 1 \]

### Step 2: Substitute the line equation into the parabola equation
To find the points of intersection, substitute \( y = mx + 1 \) into the parabola equation:
\[ (mx + 1)^2 = 4x \]

### Step 3: Expand the equation
Expanding the left-hand side:
\[ m^2x^2 + 2mx + 1 = 4x \]

### Step 4: Rearrange the equation
Rearranging gives us a quadratic equation in terms of \( x \):
\[ m^2x^2 + (2m - 4)x + 1 = 0 \]

### Step 5: Use the condition for tangency
For the line to be a 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 \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Here, \( a = m^2 \), \( b = 2m - 4 \), and \( c = 1 \). Thus, the discriminant is:
\[ D = (2m - 4)^2 - 4(m^2)(1) \]

### Step 6: Set the discriminant to zero
Setting the discriminant to zero for tangency:
\[ (2m - 4)^2 - 4m^2 = 0 \]

### Step 7: Simplify the equation
Expanding and simplifying:
\[ 4m^2 - 16m + 16 - 4m^2 = 0 \]
This simplifies to:
\[ -16m + 16 = 0 \]

### Step 8: Solve for \( m \)
Solving for \( m \):
\[ -16m = -16 \implies m = 1 \]

### Conclusion
Thus, the value of \( m \) for which the line \( y = mx + 1 \) is a tangent to the parabola \( y^2 = 4x \) is:
\[ \boxed{1} \]