If tangents PA and PB are drawn from `P(-1, 2)` to `y^(2) = 4x` then

Options

A. Equation of AB is `y = x -1`
B. Length of AB is 8
C. Length of AB is 4
D. Equation of AB is `y = x + 1 `

Answer

To find the tangents PA and PB drawn from the point P(-1, 2) to the parabola given by the equation \( y^2 = 4x \), we can follow these steps:

### Step 1: Identify the parameters of the parabola
The given parabola is \( y^2 = 4x \). In this case, we can identify \( 4a = 4 \), which gives us \( a = 1 \).

### Step 2: Write the equation of the tangent
The general equation of the tangent to the parabola \( y^2 = 4ax \) from a point \( (x_1, y_1) \) is given by:
\[ yy_1 = 2a(x + x_1) \]
Here, \( (x_1, y_1) = (-1, 2) \) and \( a = 1 \).

### Step 3: Substitute the values into the tangent equation
Substituting \( x_1 = -1 \), \( y_1 = 2 \), and \( a = 1 \) into the tangent equation:
\[ y \cdot 2 = 2 \cdot 1 \cdot (x - 1) \]
This simplifies to:
\[ 2y = 2(x - 1) \]
or
\[ y = x - 1 \]

### Step 4: Find the points of tangency
To find the points where the tangents touch the parabola, we substitute \( y = x - 1 \) into the parabola's equation \( y^2 = 4x \):
\[ (x - 1)^2 = 4x \]
Expanding this gives:
\[ x^2 - 2x + 1 = 4x \]
Rearranging the equation:
\[ x^2 - 6x + 1 = 0 \]

### Step 5: Solve the quadratic equation
Now, we can solve the quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -6, c = 1 \):
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ x = \frac{6 \pm \sqrt{36 - 4}}{2} \]
\[ x = \frac{6 \pm \sqrt{32}}{2} \]
\[ x = \frac{6 \pm 4\sqrt{2}}{2} \]
\[ x = 3 \pm 2\sqrt{2} \]

### Step 6: Find the corresponding y-values
Now, we can find the corresponding y-values using \( y = x - 1 \):
1. For \( x = 3 + 2\sqrt{2} \):
\[ y = (3 + 2\sqrt{2}) - 1 = 2 + 2\sqrt{2} \]
2. For \( x = 3 - 2\sqrt{2} \):
\[ y = (3 - 2\sqrt{2}) - 1 = 2 - 2\sqrt{2} \]

### Step 7: Write the points of tangency
The points of tangency are:
1. \( (3 + 2\sqrt{2}, 2 + 2\sqrt{2}) \)
2. \( (3 - 2\sqrt{2}, 2 - 2\sqrt{2}) \)

### Summary of the solution:
The tangents from the point P(-1, 2) to the parabola \( y^2 = 4x \) can be expressed as the line \( y = x - 1 \), and they touch the parabola at the points \( (3 + 2\sqrt{2}, 2 + 2\sqrt{2}) \) and \( (3 - 2\sqrt{2}, 2 - 2\sqrt{2}) \).