The angle between the tangents to the curve `y^(2)=2ax` at the point where `x=(a)/(2)`, is

Options

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

Answer

To find the angle between the tangents to the curve \( y^2 = 2ax \) at the point where \( x = \frac{a}{2} \), we can follow these steps:

### Step 1: Identify the points on the curve
Given the curve \( y^2 = 2ax \), we need to find the corresponding \( y \) values when \( x = \frac{a}{2} \).

\[ y^2 = 2a \left(\frac{a}{2}\right) = a^2 \]

Thus, the possible values of \( y \) are:

\[ y = \pm a \]

This gives us two points on the curve:
1. \( P_1 = \left(\frac{a}{2}, a\right) \)
2. \( P_2 = \left(\frac{a}{2}, -a\right) \)

### Step 2: Find the slope of the tangents at these points
To find the slope of the tangents, we differentiate the curve implicitly:

\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(2ax) \]
\[ 2y \frac{dy}{dx} = 2a \]

From this, we can solve for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{2a}{2y} = \frac{a}{y} \]

Now, we calculate the slopes at the two points:

1. **At point \( P_1 = \left(\frac{a}{2}, a\right) \)**:
\[ m_1 = \frac{a}{a} = 1 \]

2. **At point \( P_2 = \left(\frac{a}{2}, -a\right) \)**:
\[ m_2 = \frac{a}{-a} = -1 \]

### Step 3: Find the angle between the tangents
The angle \( \theta \) between two lines with slopes \( m_1 \) and \( m_2 \) can be calculated using the formula:

\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

Substituting the values of \( m_1 \) and \( m_2 \):

\[ \tan(\theta) = \left| \frac{1 - (-1)}{1 + (1)(-1)} \right| = \left| \frac{2}{0} \right| \]

Since the denominator is zero, this indicates that the angle \( \theta \) is \( 90^\circ \) or \( \frac{\pi}{2} \) radians.

### Final Answer
Thus, the angle between the tangents to the curve at the given point is:

\[ \theta = \frac{\pi}{2} \]