From an external point `P ,` a pair of tangents is drawn to the parabola `y^2=4xdot` If `theta_1a n dtheta_2` are the inclinations of these tangents with the x-axis such that `theta_1+theta_2=pi/4` , then find the locus of `Pdot`

Explanation

To solve the problem, we need to find the locus of the point \( P(h, k) \) from which two tangents are drawn to the parabola \( y^2 = 4ax \) such that the sum of the inclinations of the tangents is \( \frac{\pi}{4} \).

### Step 1: Write the equation of the tangents
The equation of the tangents to the parabola \( y^2 = 4ax \) can be expressed as:
\[ y = mx + \frac{a}{m} \]
where \( m \) is the slope of the tangent.

### Step 2: Substitute point \( P(h, k) \)
Since the tangents pass through the point \( P(h, k) \), we substitute these coordinates into the tangent equation:
\[ k = mh + \frac{a}{m} \]
Multiplying through by \( m \) to eliminate the fraction gives:
\[ mk = m^2h + a \]
Rearranging this, we get:
\[ m^2h - mk + a = 0 \]

### Step 3: Form a quadratic equation in \( m \)
The above equation is a quadratic in \( m \):
\[ m^2h - mk + a = 0 \]
For this quadratic to have real roots (which correspond to the two tangents), the discriminant must be non-negative:
\[ D = (-k)^2 - 4ha \geq 0 \]
This simplifies to:
\[ k^2 - 4ha \geq 0 \]

### Step 4: Use the condition on angles
Given that \( \theta_1 + \theta_2 = \frac{\pi}{4} \), we can use the tangent addition formula:
\[ \tan(\theta_1 + \theta_2) = \frac{\tan \theta_1 + \tan \theta_2}{1 - \tan \theta_1 \tan \theta_2} = 1 \]
This implies:
\[ \tan \theta_1 + \tan \theta_2 = 1 - \tan \theta_1 \tan \theta_2 \]
Let \( m_1 = \tan \theta_1 \) and \( m_2 = \tan \theta_2 \). The sum of the slopes \( m_1 + m_2 = -\frac{-k}{h} = \frac{k}{h} \) and the product \( m_1 m_2 = \frac{a}{h} \).

### Step 5: Set up the equations
From the angle condition:
\[ \frac{k}{h} = 1 - \frac{a}{h} \]
Multiplying through by \( h \):
\[ k = h - a \]

### Step 6: Replace \( a \) with \( 1 \) (assuming \( a = 1 \))
If we assume \( a = 1 \) (for simplicity), we have:
\[ k = h - 1 \]

### Step 7: Rearranging to find the locus
Rearranging gives:
\[ h = x \quad \text{and} \quad k = y \]
Thus:
\[ y = x - 1 \]

### Final Result
The locus of the point \( P \) is:
\[ x - y = 1 \quad \text{or} \quad x + y = 1 \]