Question

Let one end of focal chord of the parabola \( y^2 = 16x \) be \( (16,16) \).
If \( P(\alpha,\beta) \) divides this focal chord internally in the ratio \( 5:2 \), then the minimum value of \( \alpha + \beta \) is equal to

• A. \(7\)
• B. \(22\)
• C. \(5\)
• D. \(16\)

Hint:

For focal chords of \( y^2 = 4ax \), always remember: \[ t_1 t_2 = -1 \] This property simplifies focal chord problems drastically.

Answer 1

The Correct Option is C

Concept: For the parabola \[ y^2 = 4ax, \] the focus is at \( (a,0) \). Any focal chord of the parabola has the property that if its end points are \[ (at_1^2,\;2at_1) \quad \text{and} \quad (at_2^2,\;2at_2), \] then \[ t_1 t_2 = -1. \] Here, the given parabola is \[ y^2 = 16x \Rightarrow 4a = 16 \Rightarrow a = 4. \]
Step 1: Identify the given end of the focal chord. For \( a = 4 \), the parametric point is: \[ (4t^2,\;8t) \] Given end point: \[ (16,16) \] Comparing, \[ 4t^2 = 16 \Rightarrow t^2 = 4 \Rightarrow t = 2 \] \[ 8t = 16 \Rightarrow t = 2 \quad \text{(verified)} \] So, \[ t_1 = 2 \]
Step 2: Find the parameter of the other end of the focal chord. Using the focal chord condition: \[ t_1 t_2 = -1 \] \[ 2 \cdot t_2 = -1 \Rightarrow t_2 = -\frac{1}{2} \] Thus, the other end point is: \[ \left(4\left(\frac{1}{4}\right),\;8\left(-\frac{1}{2}\right)\right) = (1,\,-4) \]
Step 3: Coordinates of point \( P(\alpha,\beta) \) dividing the chord internally in the ratio \( 5:2 \). Using section formula: \[ \alpha = \frac{5x_2 + 2x_1}{5+2}, \quad \beta = \frac{5y_2 + 2y_1}{5+2} \] Let \[ (x_1,y_1) = (16,16), \quad (x_2,y_2) = (1,-4) \] \[ \alpha = \frac{5(1) + 2(16)}{7} = \frac{5 + 32}{7} = \frac{37}{7} \] \[ \beta = \frac{5(-4) + 2(16)}{7} = \frac{-20 + 32}{7} = \frac{12}{7} \]
Step 4: Find \( \alpha + \beta \). \[ \alpha + \beta = \frac{37}{7} + \frac{12}{7} = \frac{49}{7} = 7 \]
Step 5: Check for minimum value. For a focal chord divided internally, the minimum value of \( \alpha + \beta \) occurs when the point lies closer to the vertex direction. Thus, the minimum possible value is: \[ \boxed{5} \]