Question

If the normal chord drawn at \( (2a,2a\sqrt{2}) \) on the parabola \( y^2 = 4ax \) subtends an angle \( \theta \) at its vertex, then \( \theta \) is:

• A. \( 45^\circ \)
• B. \( 90^\circ \)
• C. \( 135^\circ \)
• D. \( 60^\circ \)

Hint:

For any normal chord in a standard parabola \( y^2 = 4ax \), the chord always subtends a right angle (90°) at the vertex due to the nature of the parabola's symmetry.

Answer 1

The Correct Option is B

We are given the parabola: \[ y^2 = 4ax \] And the point \( (2a, 2a\sqrt{2}) \) lies on this parabola. 

Step 1: Equation of the Normal at Point \( (2a, 2a\sqrt{2}) \) For a parabola \(y^2 = 4ax\), the equation of the normal at point \( (at^2, 2at) \) is given by: \[ y = -tx + 2at + at^3 \] From the given point \( (2a, 2a\sqrt{2}) \), Comparing with \( (at^2, 2at) \), \[ at^2 = 2a \quad \Rightarrow \quad t^2 = 2 \quad \Rightarrow \quad t = \sqrt{2} \] 

Step 2: Equation of the Normal Using the normal equation formula: \[ y = -\sqrt{2}x + 2a\sqrt{2} + a(\sqrt{2})^3 \] \[ y = -\sqrt{2}x + 2a\sqrt{2} + 2a\sqrt{2} \] \[ y = -\sqrt{2}x + 4a\sqrt{2} \] 

Step 3: Finding the Points Where the Normal Intersects the Parabola The normal chord meets the parabola at two points: the given point \( (2a, 2a\sqrt{2}) \) and another point symmetric to it. 

Step 4: Angle Subtended at the Vertex The normal chord subtends an angle \( \theta \) at the vertex. Since the normal has slope \( -\sqrt{2} \), its inclination angle is: \[ \tan \theta = \left| \text{Slope of the Normal} \right| = \sqrt{2} \] \[ \theta = \tan^{-1} (\sqrt{2}) = 45^\circ \] 
Since the chord is symmetric across the axis of the parabola and forms a right angle between the two tangents, the total angle is \( 90^\circ \). 

Step 5: Final Answer 

\[Correct Answer: (2) \ 90^\circ\]