The chords of contact of the pairs of tangents drawn from each point on the line `2x +y=4` to the parabola `y^2=-4x` pass through the point

Options

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

Explanation

To solve the problem, we need to find the fixed point through which the chords of contact of the pairs of tangents drawn from each point on the line \(2x + y = 4\) to the parabola \(y^2 = -4x\) pass.

### Step-by-Step Solution:

1. **Identify the given line and parabola:**
- The line is given by \(2x + y = 4\).
- The parabola is given by \(y^2 = -4x\).

2. **Parameterize a point on the line:**
- We can express a point \(P\) on the line in terms of a parameter \(h\):
\[ P(h, 4 - 2h) \]
- Here, \(x = h\) and \(y = 4 - 2h\).

3. **Write the equation of the chord of contact:**
- The equation of the chord of contact for the parabola \(y^2 = -4x\) from the point \(P(x_1, y_1)\) is given by:
\[ yy_1 + 2(x + x_1) = 0 \]
- Substituting \(x_1 = h\) and \(y_1 = 4 - 2h\):
\[ y(4 - 2h) + 2(x + h) = 0 \]

4. **Simplify the equation:**
- Expanding the equation gives:
\[ 4y - 2hy + 2x + 2h = 0 \]
- Rearranging this, we have:
\[ 2x + 4y + 2h - 2hy = 0 \]

5. **Factor the equation:**
- We can factor it as:
\[ 2x + 4y + 2h(1 - y) = 0 \]
- This can be expressed in the form \(L_1 + \lambda L_2 = 0\), indicating a family of lines passing through a fixed point.

6. **Identify the fixed point:**
- From the equation \(1 - y = 0\), we find:
\[ y = 1 \]
- Substituting \(y = 1\) into the equation \(2x + 4(1) + 2h(1 - 1) = 0\) simplifies to:
\[ 2x + 4 = 0 \implies x = -2 \]

7. **Conclusion:**
- The fixed point through which all chords of contact pass is:
\[ (-2, 1) \]

### Final Answer:
The chords of contact of the pairs of tangents drawn from each point on the line \(2x + y = 4\) to the parabola \(y^2 = -4x\) pass through the point \((-2, 1)\).