Question

Given the two parabolas: \[ S = y^2 - 4ax = 0, \quad S' = y^2 + ax = 0 \] where \( P(t) \) is a point on the parabola \( S' = 0 \). If \( A \) and \( B \) are the feet of the perpendiculars from \( P \) to the coordinate axes and \( AB \) is a tangent to the parabola \( S = 0 \) at the point \( Q(t_1) \), then the value of \( t_1 \) is:

• A. \( t \)
• B. \( \frac{t}{4} \)
• C. \( \frac{3t}{4} \)
• D. \( \frac{t}{2} \)

Hint:

For problems involving tangents to parabolas, use the parametric equation to find the equation of the line. Then compare the slope with the standard tangent equation.

Answer 1

The Correct Option is D

Step 1: Find the coordinates of \( P \) 
For the parabola \( S' = y^2 + ax = 0 \), the parametric coordinates are: \[ P(t) = \left( -\frac{t^2}{a}, t \right). \] 

Step 2: Find the feet of perpendiculars \( A \) and \( B \) 
- \( A \) is the foot of the perpendicular from \( P \) to the \( x \)-axis: \[ A = \left( -\frac{t^2}{a}, 0 \right). \] - \( B \) is the foot of the perpendicular from \( P \) to the \( y \)-axis: \[ B = (0, t). \] 

Step 3: Find the equation of line \( AB \) 
The slope of line \( AB \) is: \[ m = \frac{t - 0}{0 - (-t^2/a)} = \frac{t}{t^2/a} = \frac{a}{t}. \] Equation of \( AB \): \[ y - 0 = \frac{a}{t} \left( x + \frac{t^2}{a} \right). \] \[ y = \frac{a}{t} x + \frac{t^2}{t} = \frac{a}{t} x + t. \]

 Step 4: Condition for tangency to \( S = 0 \) 
The equation of the tangent to \( S = y^2 - 4ax = 0 \) at \( Q(t_1) \) is: \[ yy_1 = 2a(x + x_1). \] Substituting \( y_1 = t_1 \) and \( x_1 = \frac{t_1^2}{4a} \): \[ yt_1 = 2a \left( x + \frac{t_1^2}{4a} \right). \] Rewriting: \[ y = \frac{2a}{t_1} x + \frac{t_1}{2}. \] Comparing slopes: \[ \frac{a}{t} = \frac{2a}{t_1}. \] \[ t_1 = \frac{2t}{2} = \frac{t}{2}. \] 

Step 5: Conclusion 
Thus, the correct answer is: \[ \mathbf{\frac{t}{2}}. \]