The angle between tangents to the parabola `y^(2)=4x` at the points where it intersects with the line x-y -1 =0 is

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(2)`

Answer

To find the angle between the tangents to the parabola \(y^2 = 4x\) at the points where it intersects with the line \(x - y - 1 = 0\), we can follow these steps:

### Step 1: Find the points of intersection
We start by substituting \(y\) from the line equation into the parabola equation. The line can be rewritten as:
\[ y = x - 1 \]
Now substitute this into the parabola 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 2: Solve the quadratic equation
We can solve the quadratic equation \(x^2 - 6x + 1 = 0\) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -6\), and \(c = 1\):
\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{6 \pm \sqrt{36 - 4}}{2} = \frac{6 \pm \sqrt{32}}{2} = \frac{6 \pm 4\sqrt{2}}{2} = 3 \pm 2\sqrt{2} \]

### Step 3: Find corresponding \(y\) values
Now, we 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 \(x = 3 - 2\sqrt{2}\):
\[ y = (3 - 2\sqrt{2}) - 1 = 2 - 2\sqrt{2} \]

Thus, the points of intersection are:
\[ (3 + 2\sqrt{2}, 2 + 2\sqrt{2}) \quad \text{and} \quad (3 - 2\sqrt{2}, 2 - 2\sqrt{2}) \]

### Step 4: Find the slopes of the tangents
The slope of the tangent to the parabola \(y^2 = 4x\) at any point \((x_0, y_0)\) is given by:
\[ m = \frac{y_0}{2} \]
Calculating the slopes at the points of intersection:
1. At \((3 + 2\sqrt{2}, 2 + 2\sqrt{2})\):
\[ m_1 = \frac{2 + 2\sqrt{2}}{2} = 1 + \sqrt{2} \]
2. At \((3 - 2\sqrt{2}, 2 - 2\sqrt{2})\):
\[ m_2 = \frac{2 - 2\sqrt{2}}{2} = 1 - \sqrt{2} \]

### Step 5: Find the angle between the tangents
The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Calculating \(m_1 - m_2\) and \(1 + m_1 m_2\):
\[ m_1 - m_2 = (1 + \sqrt{2}) - (1 - \sqrt{2}) = 2\sqrt{2} \]
\[ m_1 m_2 = (1 + \sqrt{2})(1 - \sqrt{2}) = 1 - 2 = -1 \]
Thus,
\[ 1 + m_1 m_2 = 1 - 1 = 0 \]
Since the denominator is zero, this indicates that the tangents are perpendicular.

### Conclusion
Therefore, the angle between the tangents is:
\[ \theta = 90^\circ = \frac{\pi}{2} \]