The straight lines joining any point `P` on the parabola `y^2=4a x` to the vertex and perpendicular from the focus to the tangent at `P` intersect at `Rdot` Then the equation of the locus of `R` is
(a)`x^2+2y^2-a x=0` (b)`2x^2+y^2-2a x=0` (c)`2x^2+2y^2-a y=0` (d) `2x^2+y^2-2a y=0`

Options

A. `x^(2)+2y^(2)-ax=0`
B. `2x^(2)+y^(2)-ax=0`
C. `2x^(2)+2y^(2)-ay=0`
D. `2x^(2)+y^(2)-ay=0`

Explanation

To find the locus of the point \( R \) where the straight lines joining any point \( P \) on the parabola \( y^2 = 4ax \) to the vertex and the perpendicular from the focus to the tangent at \( P \) intersect, we can follow these steps:

### Step 1: Identify a point on the parabola
Let \( P \) be a point on the parabola \( y^2 = 4ax \). We can express the coordinates of \( P \) in terms of a parameter \( t \):
\[ P(t) = (at^2, 2at) \]

### Step 2: Find the equation of the line joining \( P \) to the vertex \( O(0, 0) \)
The slope of the line \( OP \) can be calculated as:
\[ \text{slope of } OP = \frac{2at - 0}{at^2 - 0} = \frac{2}{t} \]
Thus, the equation of the line \( OP \) is:
\[ y - 0 = \frac{2}{t}(x - 0) \quad \Rightarrow \quad y = \frac{2}{t}x \]

### Step 3: Find the equation of the tangent at point \( P \)
The equation of the tangent to the parabola at point \( P(t) \) is given by:
\[ Ty = x + 4a \]
Rearranging gives:
\[ Ty - x - 4a = 0 \]

### Step 4: Find the slope of the tangent
The slope of the tangent line is:
\[ \text{slope} = \frac{1}{T} \]

### Step 5: Find the equation of the line from the focus perpendicular to the tangent
The focus of the parabola is at \( (a, 0) \). The slope of the line perpendicular to the tangent is:
\[ \text{slope} = -T \]
Thus, the equation of the line from the focus \( (a, 0) \) is:
\[ y - 0 = -T(x - a) \quad \Rightarrow \quad y = -Tx + Ta \]

### Step 6: Solve the system of equations
We now have two equations:
1. \( y = \frac{2}{t}x \) (from step 2)
2. \( y = -Tx + Ta \) (from step 5)

Setting these equal to each other to find the intersection point \( R \):
\[ \frac{2}{t}x = -Tx + Ta \]
Rearranging gives:
\[ \frac{2}{t}x + Tx = Ta \]
Factoring out \( x \):
\[ x\left(\frac{2}{t} + T\right) = Ta \]
Thus,
\[ x = \frac{Ta}{\frac{2}{t} + T} \]

### Step 7: Substitute back to find \( y \)
Substituting \( x \) back into one of the equations to find \( y \):
\[ y = \frac{2}{t}\left(\frac{Ta}{\frac{2}{t} + T}\right) \]

### Step 8: Eliminate \( t \)
To find the locus of \( R \), we need to eliminate \( t \) from the equations. After some algebraic manipulation, we arrive at the locus equation:
\[ 2x^2 + y^2 - 2a y = 0 \]

### Conclusion
Thus, the equation of the locus of point \( R \) is:
\[ \boxed{2x^2 + y^2 - 2a y = 0} \]