If two tanents drawn from a point P to the parabola `y^2 = 4x` are at right angles, then the locus of P is

Options

A. `2x+1=0`
B. `x=-1 `
C. `2x – 1=0`
D. `x=1`

Answer

To find the locus of the point \( P(h, k) \) from which two tangents drawn to the parabola \( y^2 = 4x \) are at right angles, we can follow these steps:

### Step 1: Write the equation of the parabola and the point P
The equation of the parabola is given as:
\[ y^2 = 4x \]
Let the point \( P \) be \( (h, k) \).

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

### Step 3: Substitute point P into the tangent equation
Since the tangent passes through the point \( P(h, k) \), we substitute \( (h, k) \) into the tangent equation:
\[ k = mh + \frac{1}{m} \]

### Step 4: Rearrange the equation
Rearranging the equation gives:
\[ km = mh^2 + 1 \]
This can be rewritten as:
\[ mh^2 - km + 1 = 0 \]

### Step 5: Identify the roots of the quadratic equation
Let \( m_1 \) and \( m_2 \) be the slopes of the two tangents. From the quadratic equation, we know:
- The sum of the roots \( m_1 + m_2 = \frac{k}{h} \)
- The product of the roots \( m_1 m_2 = \frac{1}{h} \)

### Step 6: Use the condition for perpendicular tangents
For the tangents to be perpendicular, the product of the slopes must equal -1:
\[ m_1 m_2 = -1 \]
Setting the two expressions for the product of the slopes equal gives:
\[ \frac{1}{h} = -1 \]

### Step 7: Solve for h
From the equation \( \frac{1}{h} = -1 \), we can solve for \( h \):
\[ h = -1 \]

### Step 8: Write the locus of point P
Since \( h \) is constant at -1, the locus of point \( P \) can be expressed as:
\[ x = -1 \]

### Final Answer
Thus, the locus of the point \( P \) is:
\[ \text{Locus: } x + 1 = 0 \]