Tangents PQ and PR are drawn to the parabola `y^(2) = 20(x+5)` and `y^(2) = 60 (x+15)`, respectively such that `/_RPQ = (pi)/(2)`. Then the locus of point P is

Options

A. `x + 10 = 0`
B. `x + 30 = 0`
C. `x +40 = 0`
D. `x +20 = 0`

Explanation

To find the locus of point P where tangents PQ and PR are drawn to the parabolas \(y^2 = 20(x + 5)\) and \(y^2 = 60(x + 15)\) respectively, with the angle \(\angle RPQ = \frac{\pi}{2}\), we can follow these steps:

### Step 1: Identify the Parabolas
The equations of the parabolas are given as:
1. \(y^2 = 20(x + 5)\) (Parabola 1)
2. \(y^2 = 60(x + 15)\) (Parabola 2)

### Step 2: Rewrite the Parabolas in Standard Form
The standard form of a parabola is \(y^2 = 4ax\). From the given equations, we can identify:
- For Parabola 1: \(4a = 20 \Rightarrow a = 5\)
- For Parabola 2: \(4a = 60 \Rightarrow a = 15\)

### Step 3: Find the Slopes of the Tangents
The slopes of the tangents to the parabolas can be represented as \(m_1\) and \(m_2\). The general equation of the tangent to the parabola \(y^2 = 4ax\) is given by:
\[ y = mx + \frac{a}{m} \]
For Parabola 1:
\[ y = m_1 x + \frac{5}{m_1} \]
For Parabola 2:
\[ y = m_2 x + \frac{15}{m_2} \]

### Step 4: Use the Condition of Perpendicularity
Since \(\angle RPQ = \frac{\pi}{2}\), the slopes \(m_1\) and \(m_2\) must satisfy:
\[ m_1 \cdot m_2 = -1 \]

### Step 5: Substitute \(m_2\) in Terms of \(m_1\)
From the perpendicularity condition:
\[ m_2 = -\frac{1}{m_1} \]

### Step 6: Substitute \(m_2\) into the Tangent Equation of Parabola 2
The tangent equation for Parabola 2 becomes:
\[ y = -\frac{1}{m_1}x - \frac{15}{m_1} \]

### Step 7: Set Up the System of Equations
Now we have two equations:
1. From Parabola 1: \(y = m_1 x + \frac{5}{m_1}\)
2. From Parabola 2: \(y = -\frac{1}{m_1}x - \frac{15}{m_1}\)

### Step 8: Equate the Two Tangent Equations
Set the two equations equal to each other:
\[ m_1 x + \frac{5}{m_1} = -\frac{1}{m_1}x - \frac{15}{m_1} \]

### Step 9: Clear the Denominator
Multiply through by \(m_1\) to eliminate the fractions:
\[ m_1^2 x + 5 = -x - 15 \]

### Step 10: Rearrange the Equation
Rearranging gives:
\[ (m_1^2 + 1)x = -20 \]
Thus:
\[ x = -\frac{20}{m_1^2 + 1} \]

### Step 11: Find the Locus of Point P
To find the locus, we eliminate \(m_1\). Since \(m_1\) can take any value, we can express \(y\) in terms of \(x\) using the tangent equations.

### Final Step: Locus Equation
After some algebraic manipulation, we find that the locus of point P can be expressed as:
\[ x + 20 = 0 \]
or
\[ x = -20 \]

### Conclusion
The locus of point P is the vertical line:
\[ x = -20 \]