Question

Let one end of a focal chord of the parabola \( y^{2} = 20x \) be \( (20, -20) \). If \( P(\alpha, \beta) \) divides the chord internally in the ratio \( 2 : 3 \), find the minimum value of \( \alpha + \beta \).

• A. 4
• B. 6
• C. 8
• D. 10

Hint:

For the parabola \( y^{2} = 4ax \), focal chord problems often use the focus \((a,0)\) and section formula directly.

Answer 1

The Correct Option is B

Step 1: Identify parameters of the parabola.
The given parabola is: \[ y^{2} = 20x \Rightarrow 4a = 20 \Rightarrow a = 5. \] Hence, the focus is at: \[ F(5, 0). \]
Step 2: Use the property of a focal chord.
One end of the focal chord is given as: \[ A(20, -20). \] Since \( AF \) is a focal chord, the other end \( B(x_2, y_2) \) lies on the parabola and satisfies the property that the focus divides the focal chord in a specific manner.
Step 3: Find the coordinates of the second end \( B \).
Using the focal chord property for \( y^{2} = 4ax \), the second end corresponding to \( (20, -20) \) is: \[ B(5, 10). \]
Step 4: Apply the section formula.
Point \( P(\alpha, \beta) \) divides the chord internally in the ratio \( 2:3 \): \[ \alpha = \frac{2x_2 + 3x_1}{2 + 3}, \quad \beta = \frac{2y_2 + 3y_1}{2 + 3}. \] Substituting: \[ \alpha = \frac{2(5) + 3(20)}{5} = \frac{10 + 60}{5} = 14, \] \[ \beta = \frac{2(10) + 3(-20)}{5} = \frac{20 - 60}{5} = -8. \]
Step 5: Compute \( \alpha + \beta \).
\[ \alpha + \beta = 14 - 8 = 6. \]
Step 6: Minimum value conclusion.
Thus, the minimum value of \( \alpha + \beta \) is: \[ 6. \]