Question

Let one end of a 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. 5
• B. 7
• C. 16
• D. 22

Hint:

Remember the parameter relation for focal chords: \( t_1 t_2 = -1 \). This allows you to find the coordinates of the other end immediately.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A focal chord of \( y^2 = 4ax \) passes through the focus \( (a, 0) \). If one end is \( (at_1^2, 2at_1) \), the other end is \( (at_2^2, 2at_2) \) where \( t_1 t_2 = -1 \).

Step 2: Detailed Explanation:
For \( y^2 = 16x \), \( a = 4 \).
Let one end be \( A(16, 16) \). Then \( 2at_1 = 16 \implies 8t_1 = 16 \implies t_1 = 2 \).
The other end \( B \) satisfies \( t_2 = -1/t_1 = -1/2 \).
Coordinates of \( B \):
\[ x = a(t_2)^2 = 4(-1/2)^2 = 1 \]
\[ y = 2a(t_2) = 2(4)(-1/2) = -4 \]
So, \( B = (1, -4) \).
Point \( P(\alpha, \beta) \) divides the chord \( AB \) in the ratio \( 5 : 2 \).
There are two possible cases for the internal division order:
Case 1: Ratio is \( 5 : 2 \) from \( A \) to \( B \).
\[ \alpha = \frac{5(1) + 2(16)}{5+2} = \frac{37}{7}, \quad \beta = \frac{5(-4) + 2(16)}{5+2} = \frac{12}{7} \implies \alpha + \beta = \frac{49}{7} = 7 \]
Case 2: Ratio is \( 5 : 2 \) from \( B \) to \( A \).
\[ \alpha = \frac{5(16) + 2(1)}{7} = \frac{82}{7}, \quad \beta = \frac{5(16) + 2(-4)}{7} = \frac{72}{7} \implies \alpha + \beta = \frac{154}{7} = 22 \]
The minimum value is 7.

Step 3: Final Answer:
The minimum value is 7.