If a tangent to the parabola `y^(2) = 4ax` meets the x-axis in T and the tangent at the Vertex A in P and the rectangle TAPQ is completed then locus of Q is

Options

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

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript and derive the locus of point Q.

### Step-by-Step Solution:

1. **Identify Points**:
- Let point Q be represented as \( Q(H, K) \).
- Since point T is on the x-axis, its coordinates will be \( T(H, 0) \).
- The tangent at the vertex A of the parabola \( y^2 = 4ax \) will be a horizontal line, so let point P be \( P(0, K) \).

2. **Equation of Tangent**:
- The equation of the tangent to the parabola \( y^2 = 4ax \) can be expressed as:
\[ y = mx + \frac{a}{m} \]
- Here, \( m \) is the slope of the tangent.

3. **Condition of Tangency**:
- The condition of tangency gives us:
\[ \frac{a}{m} = -Hm \]
- Rearranging this gives:
\[ a = -Hm^2 \]
- This can be labeled as Equation (1).

4. **Another Condition**:
- From the point P, we also have:
\[ \frac{a}{m} = K \]
- Rearranging this gives:
\[ m = \frac{a}{K} \]
- This can be labeled as Equation (2).

5. **Substituting Values**:
- Substitute Equation (2) into Equation (1):
\[ a = -H\left(\frac{a}{K}\right)^2 \]
- Simplifying this gives:
\[ a = -\frac{a^2 H}{K^2} \]
- Canceling \( a \) (assuming \( a \neq 0 \)):
\[ 1 = -\frac{a H}{K^2} \]
- Rearranging gives:
\[ aH = -K^2 \]

6. **Locus of Q**:
- This implies:
\[ K^2 + aH = 0 \]
- Replacing \( H \) with \( x \) and \( K \) with \( y \):
\[ y^2 + ax = 0 \]
- Thus, the locus of point Q is:
\[ ax + y^2 = 0 \]

### Final Answer:
The locus of point Q is given by the equation:
\[ ax + y^2 = 0 \]