A variable chord PQ of the parabola `y^2 = 4ax` subtends a right angle at the vertex. The locus of the points of intersection of the normals at P and Q is the parabola

Options

A. `y^2 = 4a(x-2a)`
B. `y^2 = 16a(x-6a)`
C. `y^2 = 8a(x-4a)`
D. none

Answer

To solve the problem, we need to find the locus of the points of intersection of the normals at points P and Q on the parabola \( y^2 = 4ax \), where the chord PQ subtends a right angle at the vertex.

### Step-by-Step Solution:

1. **Identify Points P and Q on the Parabola:**
The points \( P \) and \( Q \) on the parabola \( y^2 = 4ax \) can be represented in parametric form as:
\[ P(t_1) = (at_1^2, 2at_1) \]
\[ Q(t_2) = (at_2^2, 2at_2) \]

2. **Condition for Right Angle:**
Since the chord PQ subtends a right angle at the vertex, we have the condition:
\[ t_1 \cdot t_2 = -4 \]
This implies that \( t_2 = -\frac{4}{t_1} \).

3. **Equations of Normals at P and Q:**
The equation of the normal at point \( P(t_1) \) is given by:
\[ y = -x t_1 + 2at_1 + at_1^3 \]
The equation of the normal at point \( Q(t_2) \) is:
\[ y = -x t_2 + 2at_2 + at_2^3 \]

4. **Finding the Intersection of Normals:**
To find the intersection of these two normals, we set the two equations equal to each other:
\[ -x t_1 + 2at_1 + at_1^3 = -x t_2 + 2at_2 + at_2^3 \]
Rearranging gives:
\[ (t_2 - t_1)x = 2a(t_2 - t_1) + (at_2^3 - at_1^3) \]
If \( t_2 \neq t_1 \), we can divide both sides by \( t_2 - t_1 \):
\[ x = 2a + a \frac{t_2^3 - t_1^3}{t_2 - t_1} \]

5. **Simplifying the Expression:**
Using the identity \( t_2^3 - t_1^3 = (t_2 - t_1)(t_2^2 + t_1t_2 + t_1^2) \), we can simplify:
\[ x = 2a + a(t_2^2 + t_1t_2 + t_1^2) \]

6. **Substituting \( t_2 \):**
Substitute \( t_2 = -\frac{4}{t_1} \) into the expression for \( x \):
\[ x = 2a + a\left(\left(-\frac{4}{t_1}\right)^2 + t_1\left(-\frac{4}{t_1}\right) + t_1^2\right) \]
Simplifying this gives:
\[ x = 2a + a\left(\frac{16}{t_1^2} - 4 + t_1^2\right) \]

7. **Finding y-coordinate:**
Substitute \( t_1 \) and \( t_2 \) into one of the normal equations to find \( y \):
\[ y = -x t_1 + 2at_1 + at_1^3 \]
After substituting \( x \) and simplifying, we will find \( y \) in terms of \( x \).

8. **Final Locus Equation:**
After performing the necessary algebraic manipulations, we arrive at the locus equation:
\[ y^2 = 16a(x - 6a) \]
This is the equation of a parabola.