The angle between tangents to the parabola `y^2 = 4ax` at the point where it intersects with the line `x - y -a=0`

Options

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

Answer

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

### Step 1: Find the points of intersection
First, we need to find the points where the parabola intersects the line. We can substitute \( y \) from the line equation into the parabola equation.

The line equation can be rearranged to:
\[ y = x - a \]

Substituting this into the parabola equation \( y^2 = 4ax \):
\[ (x - a)^2 = 4ax \]
Expanding this gives:
\[ x^2 - 2ax + a^2 = 4ax \]
Rearranging the equation yields:
\[ x^2 - 6ax + a^2 = 0 \]

### Step 2: Solve the quadratic equation
Now we can solve the quadratic equation \( x^2 - 6ax + a^2 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -6a, c = a^2 \):
\[ x = \frac{6a \pm \sqrt{(-6a)^2 - 4 \cdot 1 \cdot a^2}}{2 \cdot 1} \]
Calculating the discriminant:
\[ x = \frac{6a \pm \sqrt{36a^2 - 4a^2}}{2} \]
\[ x = \frac{6a \pm \sqrt{32a^2}}{2} \]
\[ x = \frac{6a \pm 4a\sqrt{2}}{2} \]
\[ x = 3a \pm 2a\sqrt{2} \]

### Step 3: Find the corresponding y-coordinates
Now we can find the corresponding \( y \) values using \( y = x - a \):
1. For \( x_1 = 3a + 2a\sqrt{2} \):
\[ y_1 = (3a + 2a\sqrt{2}) - a = 2a + 2a\sqrt{2} = 2a(1 + \sqrt{2}) \]
2. For \( x_2 = 3a - 2a\sqrt{2} \):
\[ y_2 = (3a - 2a\sqrt{2}) - a = 2a - 2a\sqrt{2} = 2a(1 - \sqrt{2}) \]

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

### Step 5: Find the angle between the tangents
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) can be found using the formula:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Calculating \( m_1 - m_2 \) and \( 1 + m_1 m_2 \):
\[ m_1 - m_2 = \frac{1}{1 + \sqrt{2}} - \frac{1}{1 - \sqrt{2}} = \frac{(1 - \sqrt{2}) - (1 + \sqrt{2})}{(1 + \sqrt{2})(1 - \sqrt{2})} = \frac{-2\sqrt{2}}{-1} = 2\sqrt{2} \]
\[ m_1 m_2 = \frac{1}{(1 + \sqrt{2})(1 - \sqrt{2})} = \frac{1}{-1} = -1 \]
Thus,
\[ 1 + m_1 m_2 = 1 - 1 = 0 \]

Since \( 1 + m_1 m_2 = 0 \), the tangents are perpendicular, and the angle between them is \( 90^\circ \).

### Final Answer
The angle between the tangents to the parabola at the points of intersection is \( 90^\circ \).

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