Tangents are drawn from the point (-1, 2) to the parabola `y^2 =4x` The area of the triangle for tangents and their chord of contact is

Answer

To solve the problem of finding the area of the triangle formed by the tangents drawn from the point (-1, 2) to the parabola \(y^2 = 4x\) and their chord of contact, we can follow these steps:

### Step 1: Find the Equation of the Chord of Contact
The chord of contact from a point \((x_1, y_1)\) to the parabola \(y^2 = 4x\) is given by the equation:
\[ yy_1 = 2(x + x_1) \]
Substituting \(x_1 = -1\) and \(y_1 = 2\):
\[ y \cdot 2 = 2(x - 1) \]
Simplifying this gives:
\[ y = x - 1 \]

### Step 2: Find the Points of Tangency
To find the points of tangency, we need to solve the system formed by the parabola and the chord of contact. We substitute \(y = x - 1\) into the parabola equation \(y^2 = 4x\):
\[ (x - 1)^2 = 4x \]
Expanding and rearranging:
\[ x^2 - 2x + 1 = 4x \implies x^2 - 6x + 1 = 0 \]
Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \]
Thus, the \(x\)-coordinates of the points of tangency are \(3 + 2\sqrt{2}\) and \(3 - 2\sqrt{2}\).

### Step 3: Find the Corresponding \(y\)-Coordinates
Now we find the corresponding \(y\)-coordinates using \(y = x - 1\):
1. For \(x = 3 + 2\sqrt{2}\):
\[ y = (3 + 2\sqrt{2}) - 1 = 2 + 2\sqrt{2} \]
2. For \(x = 3 - 2\sqrt{2}\):
\[ y = (3 - 2\sqrt{2}) - 1 = 2 - 2\sqrt{2} \]
Thus, the points of tangency are:
- \(B(3 + 2\sqrt{2}, 2 + 2\sqrt{2})\)
- \(C(3 - 2\sqrt{2}, 2 - 2\sqrt{2})\)

### Step 4: Calculate the Length of \(BC\)
To find the length \(BC\):
\[ BC = \sqrt{(3 + 2\sqrt{2} - (3 - 2\sqrt{2}))^2 + ((2 + 2\sqrt{2}) - (2 - 2\sqrt{2}))^2} \]
This simplifies to:
\[ BC = \sqrt{(4\sqrt{2})^2 + (4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8 \]

### Step 5: Calculate the Height from Point A to Line BC
The height is the perpendicular distance from point \(A(-1, 2)\) to the line \(y = x - 1\). The line can be rewritten as:
\[ x - y - 1 = 0 \]
Using the formula for the distance from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\):
\[ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Substituting \(A = 1\), \(B = -1\), \(C = -1\), and \((x_0, y_0) = (-1, 2)\):
\[ \text{Distance} = \frac{|1(-1) + (-1)(2) - 1|}{\sqrt{1^2 + (-1)^2}} = \frac{|-1 - 2 - 1|}{\sqrt{2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2} \]

### Step 6: Calculate the Area of Triangle ABC
The area \(A\) of triangle \(ABC\) is given by:
\[ A = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 8 \times 2\sqrt{2} = 8\sqrt{2} \]

### Final Answer
The area of the triangle formed by the tangents and their chord of contact is:
\[ \boxed{8\sqrt{2}} \text{ square units} \]