The line `x-y+2=0` touches the parabola `y^2 = 8x` at the point (A) `(2, -4)` (B) `(1, 2sqrt(2))` (C) `(4, -4 sqrt(2)` (D) `(2, 4)`

Options

A. `(2,-4)`
B. `(1,2sqrt(2))`
C. `(4,4sqrt(2))`
D. `(2,4)`

Answer

To determine the point at which the line \( x - y + 2 = 0 \) touches the parabola \( y^2 = 8x \), we can follow these steps:

### Step 1: Rewrite the line equation
The given line equation is:
\[ x - y + 2 = 0 \]
We can rearrange this to express \( y \) in terms of \( x \):
\[ y = x + 2 \]

### Step 2: Substitute \( y \) in the parabola equation
The equation of the parabola is:
\[ y^2 = 8x \]
Now, substitute \( y = x + 2 \) into the parabola equation:
\[ (x + 2)^2 = 8x \]

### Step 3: Expand and simplify the equation
Expanding the left side:
\[ x^2 + 4x + 4 = 8x \]
Now, rearranging this equation gives:
\[ x^2 + 4x + 4 - 8x = 0 \]
Simplifying further:
\[ x^2 - 4x + 4 = 0 \]

### Step 4: Factor the quadratic equation
The quadratic can be factored as:
\[ (x - 2)^2 = 0 \]
This implies:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

### Step 5: Find the corresponding \( y \) value
Now that we have \( x = 2 \), we can find \( y \) using the equation \( y = x + 2 \):
\[ y = 2 + 2 = 4 \]

### Step 6: Write the point of tangency
Thus, the point at which the line touches the parabola is:
\[ (2, 4) \]

### Conclusion
The required answer is option (D) \( (2, 4) \).

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