Question

The normal drawn at a point \( (2, -4) \) on the parabola \( y^2 = 8x \) cuts again the same parabola at \( (\alpha, \beta) \). Then \( \alpha + \beta \) is:

• A. \( 8 \)
• B. \( 16 \)
• C. \( 24 \)
• D. \( 30 \)

Hint:

For a parabola \( y^2 = 4ax \), the normal at \( (x_1, y_1) \) is: \( y - y_1 = -\frac{y_1}{2a} (x - x_1). \)

Answer 1

The Correct Option is D

Step 1: Equation of the normal For a parabola \( y^2 = 4ax \), the normal at \( (x_1, y_1) \) is given by: \( y - y_1 = -\frac{y_1}{2a} (x - x_1). \) For \( y^2 = 8x \) (\( 4a = 8 \Rightarrow a = 2 \)), at \( (2, -4) \): \( y + 4 = -\frac{-4}{4} (x - 2). \) Simplifying: \( y + 4 = x - 2. \) \( x - y = 6. \) 

Step 2: Finding the second intersection Substituting \( x = y + 6 \) in \( y^2 = 8x \): \( y^2 = 8(y + 6). \) \( y^2 - 8y - 48 = 0. \) 

Solving for \( y \), \( y = \frac{8 \pm \sqrt{64 + 192}}{2} = \frac{8 \pm 16}{2}. \) So \( y = 12 \) or \( y = -4 \). Taking the second intersection, \( \beta = 12 \). \( \alpha = \frac{12^2}{8} = 18. \) \( \alpha + \beta = 18 + 12 = 30. \)