If the line px + qy =1m is a tangent to the parabola `y^(2)` =4ax, then

Options

A. `p^2-aq=0`
B. `p+aq^2=0`
C. `p-aq^2=0`
D. `p-2aq^2=0`

Answer

To solve the problem, we need to determine the relationship between the coefficients of the line \( px + qy = 1 \) and the parabola \( y^2 = 4ax \) when the line is a tangent to the parabola.

### Step-by-Step Solution:

1. **Equation of the Line**:
The line is given by the equation:
\[ px + qy = 1 \]
We can express \( x \) in terms of \( y \):
\[ x = \frac{1 - qy}{p} \]

2. **Substituting into the Parabola**:
Substitute \( x \) into the parabola's equation \( y^2 = 4ax \):
\[ y^2 = 4a\left(\frac{1 - qy}{p}\right) \]
Simplifying this gives:
\[ y^2 = \frac{4a(1 - qy)}{p} \]

3. **Rearranging the Equation**:
Multiply both sides by \( p \) to eliminate the fraction:
\[ py^2 = 4a(1 - qy) \]
Expanding the right-hand side:
\[ py^2 = 4a - 4aqy \]
Rearranging gives:
\[ py^2 + 4aqy - 4a = 0 \]

4. **Identifying the Quadratic Form**:
This is a quadratic equation in \( y \):
\[ py^2 + 4aqy - 4a = 0 \]
For the line to be tangent to the parabola, the discriminant of this quadratic must be zero.

5. **Calculating the Discriminant**:
The discriminant \( D \) of the quadratic equation \( Ay^2 + By + C = 0 \) is given by:
\[ D = B^2 - 4AC \]
Here, \( A = p \), \( B = 4aq \), and \( C = -4a \). Thus:
\[ D = (4aq)^2 - 4(p)(-4a) \]
Simplifying this:
\[ D = 16a^2q^2 + 16ap \]

6. **Setting the Discriminant to Zero**:
For tangency, set the discriminant equal to zero:
\[ 16a^2q^2 + 16ap = 0 \]
Dividing through by 16:
\[ a^2q^2 + ap = 0 \]

7. **Factoring the Equation**:
Factoring gives:
\[ a(aq^2 + p) = 0 \]
Since \( a \neq 0 \) for a parabola, we have:
\[ aq^2 + p = 0 \]
This implies:
\[ p + aq^2 = 0 \quad \text{or} \quad p = -aq^2 \]

### Conclusion:
Thus, the condition for the line \( px + qy = 1 \) to be a tangent to the parabola \( y^2 = 4ax \) is:
\[ p + aq^2 = 0 \]