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

To solve this problem, we begin by analyzing the given parabola \( y^2 = 4ax \) with the normal chord drawn at the point \( (2a,2a\sqrt{2}) \).

First, identify the slope of the tangent at the point \( (2a,2a\sqrt{2}) \). The formula for the slope of the tangent to the parabola \( y^2=4ax \) is \( m = \frac{y_1}{2a} \), where the point is \((x_1,y_1)\). Thus, the slope is \( \sqrt{2}/2 \).

The equation of the tangent line is: \( y - 2a\sqrt{2} = \frac{\sqrt{2}}{2}(x - 2a) \).

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope, which gives us a slope of \(-\sqrt{2}\).

The equation for the normal line is: \( y - 2a\sqrt{2} = -\sqrt{2}(x - 2a) \).

Simplifying, we get \( y = -\sqrt{2}x + (2a+2a\sqrt{2}) \).

The normal meets the parabola again at another point \((h,k)\) other than the original \( (2a,2a\sqrt{2}) \). Substituting \( y = -\sqrt{2}x + (2a+2a\sqrt{2}) \) into the parabola equation, we have:

\([-\sqrt{2}x + (2a+2a\sqrt{2})]^2 = 4ax\).

Solving this quadratic equation will give the other point of intersection. However, for checking the angle at the vertex, directly substituting isn't required.

Since the normal chord's line is calculated and must meet the parabola symmetrically, the angle subtended at the vertex by endpoints is given by:

\(\theta = \) angle between the lines joining vertex (0,0) to the chord's endpoints.

Plug and simplify the slopes, remembering the symmetry: \( m_1 \mathrm{\ is\ } \sqrt{2} \ \mathrm{and\ } m_2 \mathrm{\ is\ } -\sqrt{2} \).\

Using the formula for angle between lines, \(\tan \theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|\):

\(\tan \theta = \left|\frac{\sqrt{2} - (-\sqrt{2})}{1+(\sqrt{2})(-\sqrt{2})}\right|= \left|\frac{2\sqrt{2}}{1-2}\right| = 2\sqrt{2}\).

Thus, \(\theta = 90^\circ\), or perpendicular at the vertex.

Answer 2

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 

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