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

Options

A. 8
B. `8sqrt3`
C. `8sqrt2`
D. None of these

Answer

To find 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 \( (x_1, y_1) = (-1, 2) \):
\[ y \cdot 2 = 2(x - 1) \]
This simplifies to:
\[ 2y = 2x - 2 \implies y = x - 1 \]

### Step 2: Find the points of intersection of the chord of contact and the parabola
To find the points of intersection \( A \) and \( B \), we substitute \( y = x - 1 \) into the parabola's equation \( y^2 = 4x \):
\[ (x - 1)^2 = 4x \]
Expanding and rearranging gives:
\[ x^2 - 2x + 1 = 4x \implies x^2 - 6x + 1 = 0 \]

### Step 3: Solve the quadratic equation
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 points \( A \) and \( B \) are \( 3 + 2\sqrt{2} \) and \( 3 - 2\sqrt{2} \).

### Step 4: Find the corresponding y-coordinates
Using \( y = x - 1 \):
- For \( x = 3 + 2\sqrt{2} \), \( y = (3 + 2\sqrt{2}) - 1 = 2 + 2\sqrt{2} \)
- For \( x = 3 - 2\sqrt{2} \), \( y = (3 - 2\sqrt{2}) - 1 = 2 - 2\sqrt{2} \)

Thus, the points \( A \) and \( B \) are:
- \( A(3 + 2\sqrt{2}, 2 + 2\sqrt{2}) \)
- \( B(3 - 2\sqrt{2}, 2 - 2\sqrt{2}) \)

### Step 5: Calculate the distance \( AB \)
Using the distance formula:
\[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ AB = \sqrt{((3 - 2\sqrt{2}) - (3 + 2\sqrt{2}))^2 + ((2 - 2\sqrt{2}) - (2 + 2\sqrt{2}))^2} \]
This simplifies to:
\[ AB = \sqrt{(-4\sqrt{2})^2 + (-4\sqrt{2})^2} = \sqrt{32 + 32} = \sqrt{64} = 8 \]

### Step 6: Calculate the height from point \( P(-1, 2) \) to line \( AB \)
The equation of line \( AB \) is:
\[ y - (2 + 2\sqrt{2}) = \frac{(-4\sqrt{2})}{(-4\sqrt{2})}(x - (3 + 2\sqrt{2})) \]
This simplifies to:
\[ y = -x + 5 + 2\sqrt{2} \]
To find the height from point \( P(-1, 2) \) to this line, we use the distance from a point to a line formula:
\[ \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
Where \( A = 1, B = 1, C = -5 - 2\sqrt{2} \):
\[ \text{Distance} = \frac{|1(-1) + 1(2) - (5 + 2\sqrt{2})|}{\sqrt{1^2 + 1^2}} = \frac{|1 - 5 - 2\sqrt{2}|}{\sqrt{2}} = \frac{| -4 - 2\sqrt{2}|}{\sqrt{2}} = \frac{4 + 2\sqrt{2}}{\sqrt{2}} = 2\sqrt{2} + 2 \]

### Step 7: Calculate the area of triangle \( PAB \)
The area \( A \) of triangle \( PAB \) is given by:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 2\sqrt{2} = 8\sqrt{2} \]

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