Question

If the normal drawn at the point \( P(9, 9) \) on the parabola \( y^2 = 9x \) meets the parabola again at \( Q(a, b) \), then \( 2a + b = \text{?} \)

• A. 54
• B. \( \frac{99}{2} \)
• C. \( \frac{63}{2} \)
• D. 27 \bigskip

Hint:

To solve for the coordinates of the point where the normal intersects the parabola again, substitute \( y^2 = 9x \) into the normal equation and solve for \( y \), then find \( x \) using \( x = \frac{y^2}{9} \).

Answer 1

The Correct Option is D

We are given the parabola: \[ y^2 = 9x \] and the point \( P(9,9) \) on the parabola. We need to find where the normal at \( P \) meets the parabola again at \( Q(a,b) \) and compute: \[ 2a + b \] --- Step 1: Find Parameter \( t \) for \( P(9,9) \) For the standard parabola \( y^2 = 4ax \), we compare with \( y^2 = 9x \) to get: \[ 4a = 9 \Rightarrow a = \frac{9}{4} \] The parametric form of the parabola: \[ x = at^2, \quad y = 2at \] Substituting \( P(9,9) \): \[ 9 = \frac{9}{4} t^2 \Rightarrow t^2 = 4 \Rightarrow t = \pm 2 \] Since \( y = 2at = 9 \), solving for \( t \): \[ 2 \times \frac{9}{4} \times t = 9 \] \[ \frac{18}{4} t = 9 \] \[ \frac{9}{2} t = 9 \Rightarrow t = 2 \] Thus, \( P \) corresponds to \( t = 2 \). --- Step 2: Find Equation of the Normal at \( P \) The equation of the normal to the parabola \( y^2 = 4ax \) at the parametric point \( (at^2, 2at) \) is: \[ y - 2at = -t(x - at^2) \] Substituting \( t = 2 \) and \( a = \frac{9}{4} \): \[ y - 2 \times \frac{9}{4} \times 2 = -2(x - \frac{9}{4} \times 4) \] \[ y - 9 = -2(x - 9) \] \[ y - 9 = -2x + 18 \] \[ y = -2x + 27 \] --- Step 3: Find the Second Intersection \( Q(a,b) \) To find the second intersection, substitute \( y = -2x + 27 \) into the parabola equation: \[ (-2x + 27)^2 = 9x \] Expanding: \[ 4x^2 - 108x + 729 = 9x \] \[ 4x^2 - 117x + 729 = 0 \] Solving for \( x \) using the quadratic formula: \[ x = \frac{-(-117) \pm \sqrt{(-117)^2 - 4(4)(729)}}{2(4)} \] \[ x = \frac{117 \pm \sqrt{13689 - 11664}}{8} \] \[ x = \frac{117 \pm \sqrt{2025}}{8} \] \[ x = \frac{117 \pm 45}{8} \] \[ x = \frac{162}{8} = \frac{81}{4} \quad \text{or} \quad x = \frac{72}{8} = 9 \] Since \( P(9,9) \) is one root, the second intersection \( Q(a,b) \) corresponds to: \[ a = \frac{81}{4} \] Finding \( b \): \[ b = -2 \left( \frac{81}{4} \right) + 27 \] \[ = -\frac{162}{4} + 27 = -\frac{162}{4} + \frac{108}{4} \] \[ = \frac{-162 + 108}{4} = \frac{-54}{4} = -\frac{27}{2} \] --- Step 4: Compute \( 2a + b \) \[ 2a + b = 2 \times \frac{81}{4} + \left( -\frac{27}{2} \right) \] \[ = \frac{162}{4} - \frac{54}{4} \] \[ = \frac{108}{4} = 27 \] --- Final Answer: \[ \boxed{27} \] \bigskip