Tangents `PA` and `PB` are drawn to parabola `y^(2)=4x` from any arbitrary point `P` on the line `x+y=1`.
Then vertex of locus of midpoint of chord `AB` is

Options

A. `(-1,-2)`
B. `(3/2,1)`
C. `(-3/2,-1)`
D. `(3/2,-1)`

Explanation

To find the vertex of the locus of the midpoint of the chord \( AB \) formed by tangents drawn from a point \( P \) on the line \( x + y = 1 \) to the parabola \( y^2 = 4x \), we can follow these steps:

### Step 1: Identify the point on the line
Let the point \( P \) on the line \( x + y = 1 \) be represented as \( P(\alpha, 1 - \alpha) \).

### Step 2: Write the equation of the chord of contact
The equation of the chord of contact for the parabola \( y^2 = 4x \) from the point \( P(\alpha, 1 - \alpha) \) is given by:
\[ y(1 - \alpha) = 2(x + \alpha) \]
Rearranging this, we get:
\[ 2x - y(1 - \alpha) + 2\alpha = 0 \]
This is our equation (1).

### Step 3: Midpoint of the chord
Let \( (h, k) \) be the midpoint of the chord \( AB \). The equation of the parabola \( y^2 = 4x \) can be used to express the relationship between the coordinates of the midpoint and the tangents.

Using the formula for the chord of contact, we have:
\[ T = S_1 \]
where \( T \) is the equation of the tangent at the midpoint \( (h, k) \) and \( S_1 \) is the equation of the parabola at that point.

The equation of the tangent at point \( (h, k) \) is:
\[ ky = 2(x + h) \]
Rearranging gives us:
\[ 2x - ky + 2h = 0 \]
This is our equation (2).

### Step 4: Equate the two equations
Now we equate the coefficients of equations (1) and (2):
From equation (1):
\[ 2x - y(1 - \alpha) + 2\alpha = 0 \]
From equation (2):
\[ 2x - ky + 2h = 0 \]
By comparing coefficients, we get:
1. \( 1 - \alpha = k \)
2. \( 2\alpha = 2h \)

### Step 5: Solve for \( h \) and \( k \)
From the second equation, we have:
\[ h = \alpha \]
Substituting this into the first equation gives us:
\[ k = 1 - h \]
Thus, we can express \( k \) in terms of \( h \):
\[ k = 1 - \alpha \]

### Step 6: Find the locus
Substituting \( h \) and \( k \) into the equation gives:
\[ k = 1 - h \]
This implies:
\[ k + h = 1 \]
Now substituting \( h \) and \( k \) into the equation of the parabola gives us:
\[ k^2 = 4h \]
Substituting \( k = 1 - h \) into this equation:
\[ (1 - h)^2 = 4h \]
Expanding and rearranging gives:
\[ 1 - 2h + h^2 = 4h \]
\[ h^2 - 6h + 1 = 0 \]

### Step 7: Find the vertex
To find the vertex of the locus, we can complete the square or use the vertex formula \( h = -\frac{b}{2a} \):
\[ h = \frac{6}{2} = 3 \]
Substituting back to find \( k \):
\[ k = 1 - 3 = -2 \]
Thus, the vertex of the locus is \( (3, -2) \).

### Final Answer
The vertex of the locus of the midpoint of the chord \( AB \) is \( (3, -2) \).

---