Step 1: Understanding the Question:
Tangency occurs when a line touches a curve at exactly one point.
In coordinate geometry, there are specific algebraic conditions that relate the slope and intercept of a line to the parameters of a conic section for tangency to occur.
We are given a linear equation \( y = mx + 2 \) and a parabola \( y^2 = 8x \) and need to find the specific value of \( m \) that satisfies this condition.
Step 2: Key Formula or Approach:
1. For the standard parabola \( y^2 = 4ax \), the condition for a line \( y = mx + c \) to be a tangent is given by the formula: \( c = \frac{a}{m} \).
2. First, identify the parameter \( a \) from the parabola's equation.
3. Then, identify \( c \) from the line's equation and solve for \( m \).
Step 3: Detailed Explanation:
Identify the parameters:
The given parabola is \( y^2 = 8x \).
Comparing this with \( y^2 = 4ax \), we have \( 4a = 8 \), which implies \( a = 2 \).
The given line is \( y = mx + 2 \).
Comparing this with \( y = mx + c \), we find that the y-intercept is \( c = 2 \).
Applying the condition of tangency:
Substitute \( a = 2 \) and \( c = 2 \) into the condition \( c = \frac{a}{m} \):
\[ 2 = \frac{2}{m} \]
Solving for m:
Cross-multiplying gives:
\[ 2m = 2 \]
\[ m = 1 \]
Verification:
If \( m = 1 \), the line is \( y = x + 2 \).
Substituting into the parabola: \( (x+2)^2 = 8x \implies x^2 + 4x + 4 = 8x \implies x^2 - 4x + 4 = 0 \).
Since this is a perfect square \( (x-2)^2 = 0 \), there is only one point of intersection (\( x=2, y=4 \)), which confirms the line is a tangent.
Step 4: Final Answer:
The condition for tangency is met when \( m = 1 \).