`T P` and `T Q` are tangents to the parabola `y^2=4a x` at `Pa n dQ ,` respectively. If the chord `P Q` passes through the fixed point `(-a ,b),` then find the locus of `Tdot`

Explanation

To solve the problem, we need to find the locus of the point \( T(h, k) \) from which tangents \( TP \) and \( TQ \) are drawn to the parabola \( y^2 = 4ax \) at points \( P \) and \( Q \), respectively. The chord \( PQ \) passes through the fixed point \( (-a, b) \).

### Step-by-Step Solution:

1. **Understand the Parabola**: The given parabola is \( y^2 = 4ax \). This is a standard form of a parabola that opens to the right.

2. **Equation of the Chord of Contact**: The equation of the chord of contact from a point \( T(h, k) \) to the parabola \( y^2 = 4ax \) is given by:
\[ k y = 2a(x + h) \]
This can be rearranged to:
\[ ky = 2ax + 2ah \]

3. **Substituting the Fixed Point**: Since the chord \( PQ \) passes through the fixed point \( (-a, b) \), we substitute \( x = -a \) and \( y = b \) into the chord of contact equation:
\[ kb = 2a(-a + h) \]

4. **Simplifying the Equation**: Rearranging the above equation gives:
\[ kb = -2a^2 + 2ah \]
This can be rewritten as:
\[ 2ah = kb + 2a^2 \]

5. **Finding the Locus**: We want to express this in terms of \( h \) and \( k \) without any constants. Rearranging gives:
\[ h = \frac{kb + 2a^2}{2a} \]
To eliminate \( a \), we can express \( h \) in terms of \( k \) and \( b \):
\[ h = \frac{kb}{2} + a \]

6. **Final Locus Equation**: Rearranging gives us the locus of point \( T \):
\[ 2h - kb = 2a \]
Since \( a \) is a constant, we can express the locus in the form:
\[ 2h - kb = C \]
where \( C \) is a constant depending on the specific value of \( a \).

### Locus Equation:
Thus, the locus of point \( T \) is given by:
\[ 2h - kb = 2a \]