From the point (-1,2), tangent lines are to the parabola `y^(2)=4x`. If the area of the triangle formed by the chord of contact and the tangents is A, then the value of `A//sqrt(2)` is ___________ .

Explanation

To solve the problem step by step, we will follow the outlined approach in the video transcript.

### Step 1: Identify the parabola and point
The given parabola is \( y^2 = 4x \). The point from which the tangents are drawn is \( O(-1, 2) \).

### Step 2: Write the equation of the chord of contact
The equation for the chord of contact from a point \( (x_1, y_1) \) to the parabola \( y^2 = 4ax \) is given by:
\[ yy_1 = 2a(x + x_1) \]
Here, \( a = 1 \) (since \( 4a = 4 \)), and substituting \( (x_1, y_1) = (-1, 2) \):
\[ y \cdot 2 = 2 \cdot 1 \cdot (x - 1) \]
This simplifies to:
\[ 2y = 2x - 2 \]
Dividing by 2 gives:
\[ y = x - 1 \quad \text{(Equation 2)} \]

### Step 3: Find the points of intersection
Next, we need to find the points of intersection of the chord of contact \( y = x - 1 \) with the parabola \( y^2 = 4x \). Substituting \( y \) from Equation 2 into the parabola:
\[ (x - 1)^2 = 4x \]
Expanding this:
\[ x^2 - 2x + 1 = 4x \]
Rearranging gives:
\[ x^2 - 6x + 1 = 0 \]

### Step 4: Solve the quadratic equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1, b = -6, c = 1 \):
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} \]
This simplifies to:
\[ x = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \]
Thus, the x-coordinates of the points of intersection are \( x_1 = 3 + 2\sqrt{2} \) and \( x_2 = 3 - 2\sqrt{2} \).

### Step 5: Find corresponding y-coordinates
Using \( y = x - 1 \):
- For \( x_1 = 3 + 2\sqrt{2} \):
\[ y_1 = (3 + 2\sqrt{2}) - 1 = 2 + 2\sqrt{2} \]
- For \( x_2 = 3 - 2\sqrt{2} \):
\[ y_2 = (3 - 2\sqrt{2}) - 1 = 2 - 2\sqrt{2} \]
Thus, the points of intersection are:
\[ P(3 + 2\sqrt{2}, 2 + 2\sqrt{2}), \quad Q(3 - 2\sqrt{2}, 2 - 2\sqrt{2}) \]

### Step 6: Calculate the distance \( PQ \)
The distance \( PQ \) can be calculated using the distance formula:
\[ PQ = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \]
Calculating \( x_1 - x_2 \) and \( y_1 - y_2 \):
\[ x_1 - x_2 = (3 + 2\sqrt{2}) - (3 - 2\sqrt{2}) = 4\sqrt{2} \]
\[ y_1 - y_2 = (2 + 2\sqrt{2}) - (2 - 2\sqrt{2}) = 4\sqrt{2} \]
Thus:
\[ PQ = \sqrt{(4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8 \]

### Step 7: Find the height from point O to line PQ
The line \( PQ \) has the equation \( y = x - 1 \). The perpendicular distance from point \( O(-1, 2) \) to this line can be calculated using the formula:
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( y = x - 1 \), we can rewrite it as \( x - y - 1 = 0 \):
Here, \( A = 1, B = -1, C = -1 \):
\[ \text{Distance} = \frac{|1(-1) + (-1)(2) - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|-1 - 2 - 1|}{\sqrt{2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Step 8: Calculate the area of triangle formed
The area \( A \) of triangle formed by the chord of contact and the tangents is given by:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times PQ \times \text{Distance} \]
Substituting the values:
\[ A = \frac{1}{2} \times 8 \times 2\sqrt{2} = 8\sqrt{2} \]

### Step 9: Find \( \frac{A}{\sqrt{2}} \)
Finally, we need to find:
\[ \frac{A}{\sqrt{2}} = \frac{8\sqrt{2}}{\sqrt{2}} = 8 \]

### Final Answer
Thus, the value of \( \frac{A}{\sqrt{2}} \) is \( \boxed{8} \).