The point P on the parabola `y^(2)=4ax` for which | PR-PQ | is maximum, where R(-a, 0) and Q (0, a) are two points,

Options

A. (a, 2a)
B. (a, -2a)
C. (4a, 4a)
D. (4a, -4a)

Answer

To solve the problem, we need to find the point P on the parabola \( y^2 = 4ax \) such that the expression \( |PR - PQ| \) is maximized, where \( R(-a, 0) \) and \( Q(0, a) \) are given points.

### Step 1: Define the point P on the parabola
The point P on the parabola can be represented in parametric form as:
\[ P(at^2, 2at) \]
where \( t \) is a parameter.

### Step 2: Calculate distances PR and PQ
Next, we need to find the distances \( PR \) and \( PQ \).

1. **Distance PR**:
The distance \( PR \) from point P to point R is given by:
\[ PR = \sqrt{(at^2 - (-a))^2 + (2at - 0)^2} = \sqrt{(at^2 + a)^2 + (2at)^2} \]
Simplifying this:
\[ PR = \sqrt{(a(t^2 + 1))^2 + (2at)^2} = \sqrt{a^2(t^2 + 1)^2 + 4a^2t^2} \]
\[ PR = \sqrt{a^2(t^4 + 2t^2 + 1 + 4t^2)} = \sqrt{a^2(t^4 + 6t^2 + 1)} = a\sqrt{t^4 + 6t^2 + 1} \]

2. **Distance PQ**:
The distance \( PQ \) from point P to point Q is given by:
\[ PQ = \sqrt{(at^2 - 0)^2 + (2at - a)^2} = \sqrt{(at^2)^2 + (2at - a)^2} \]
Simplifying this:
\[ PQ = \sqrt{(at^2)^2 + (2at - a)^2} = \sqrt{a^2t^4 + (2at - a)^2} \]
\[ = \sqrt{a^2t^4 + (2at - a)(2at - a)} = \sqrt{a^2t^4 + 4a^2t^2 - 4a^2t + a^2} \]
\[ = \sqrt{a^2(t^4 + 4t^2 - 4t + 1)} = a\sqrt{t^4 + 4t^2 - 4t + 1} \]

### Step 3: Set up the equation for |PR - PQ|
Now we need to maximize \( |PR - PQ| \):
\[ |PR - PQ| = \left| a\sqrt{t^4 + 6t^2 + 1} - a\sqrt{t^4 + 4t^2 - 4t + 1} \right| \]
This simplifies to:
\[ = a \left| \sqrt{t^4 + 6t^2 + 1} - \sqrt{t^4 + 4t^2 - 4t + 1} \right| \]

### Step 4: Find the derivative and maximize
To maximize this expression, we can take the derivative of the inside function with respect to \( t \) and set it to zero:
\[ \frac{d}{dt} \left( \sqrt{t^4 + 6t^2 + 1} - \sqrt{t^4 + 4t^2 - 4t + 1} \right) = 0 \]
After calculating the derivative and simplifying, we find that the critical point occurs at \( t = 1 \).

### Step 5: Find the coordinates of point P
Substituting \( t = 1 \) back into the parametric equations for P:
\[ P(a(1^2), 2a(1)) = P(a, 2a) \]

### Conclusion
Thus, the point \( P \) on the parabola \( y^2 = 4ax \) for which \( |PR - PQ| \) is maximum is:
\[ \boxed{(a, 2a)} \]