Problem 122 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

\(\mathrm{P}\) and \(\mathrm{Q}\) are two distinct points on the parabola, \(\mathrm{y}^{2}=4 \mathrm{x}\), with parameters \(t\) and \(t_{1}\) respectively. If the normal at \(P\) passes through \(\mathrm{Q}\), then the minimum value of \(\mathrm{t}_{1}^{2}\) is : (a) 8 (b) 4 (c) 6 (d) 2

Step-by-Step Solution

Answer

The minimum value of \( t_1^2 \) is 2 (answer d).

Step 1: Identify coordinates of points P and Q

The given parabola is \( y^2 = 4x \), which can be parameterized as \( P(t) = (t^2, 2t) \) and \( Q(t_1) = (t_1^2, 2t_1) \) for parameters \( t \) and \( t_1 \) respectively.

Step 2: Use the equation of the normal at P

The equation of the normal to the parabola \( y^2 = 4x \) at point \( P(t) = (t^2, 2t) \) is given by \( y + tx = 2t + t^3 \).

Step 3: Substitute coordinates of Q into the normal equation

Substitute the coordinates of \( Q(t_1) = (t_1^2, 2t_1) \) into the normal equation to check if it lies on the normal:\[ 2t_1 + tt_1^2 = 2t + t^3 \].

Step 4: Simplify the equation

Rearrange the equation:\[ tt_1^2 - 2t_1 + t^3 - 2t = 0 \]. This is a quadratic equation in terms of \( t_1 \):\[ tt_1^2 - 2t_1 + (t^3 - 2t) = 0 \].

Step 5: Determine conditions for \( t_1^2 \)

For the quadratic equation to have real roots, the discriminant \( \, D = b^2 - 4ac \, \) must be non-negative. Here, \( a = t, \; b = -2, \; c = t^3 - 2t \). Calculate the discriminant:\[ D = (-2)^2 - 4(t)(t^3 - 2t) = 4 - 4(t^4 - 2t^2) \]\[ D = 4 - 4t^4 + 8t^2 \].

Step 6: Solve for minimum \( t_1^2 \) when \( D \geq 0 \)

Rewrite the inequality:\[ 4 - 4t^4 + 8t^2 \geq 0 \]\[ t^4 - 2t^2 - 1 \leq 0 \].Let \( x = t^2 \), then \( x^2 - 2x - 1 \leq 0 \). By solving this, we find:\[ x = \frac{2 \pm \sqrt{4 + 4}}{2} \]\[ x = 1 \pm \sqrt{2} \].The valid range for \( t^2 \) is \( 1 - \sqrt{2} \leq t^2 \leq 1 + \sqrt{2} \). To find the possible minimum of \( t_1^2 \), substitute back \( t^2 = 2 \), giving the minimum \( t_1^2 = 2 \).

Step 7: Verify the solution choice

Since \( t_1^2 = 2 \) lies within the valid range and matches the question's requirement for minimum value, this confirms our solution.

Key Concepts

Parabola

A parabola is a U-shaped curve that can open upwards, downwards, or to the sides. In the context of planar geometry, it is described using quadratic equations. For the exercise above, the specific parabola is described by the equation \( y^2 = 4x \). This is a common form when discussing parabolas that open horizontally. The points on a parabola like this can be expressed using parametric equations, which we'll discuss later.

• The equation \( y^2 = 4x \) indicates that opening of the parabola is directed towards the positive x-axis.
• The vertex of this parabola is at the origin, \( (0, 0) \), where it achieves its minimum or maximum value, depending on its orientation.
• Understanding the basic properties of a parabola helps in analyzing and solving problems related to its geometry.

A parabola is crucial in various mathematical contexts, including physics and engineering, as it describes the path followed by an object under uniform gravity.

Normal to Parabola

The normal to a parabola is a line perpendicular to the tangent of the parabola at a given point. For the parabola \( y^2 = 4x \), the normal line equation can be determined using calculus or geometry. When dealing with parameterized points like \( P(t) = (t^2, 2t) \), the equation for the normal at \( P \) becomes \( y + tx = 2t + t^3 \).

• Normals are important because they provide insights into the curve's geometry and are a foundation for complex geometric constructions.
• Normals have applications in optimization and path-planning problems, where deviations from the main path are analyzed.
• To find if another point \( Q \) lies on this normal line, its coordinates need to satisfy the normal's equation.

By examining the intersection of normals with other points or lines, we can solve various mathematical problems, as the one in the exercise.

Quadratic Discriminant

The quadratic discriminant is an expression used to determine the nature of the roots of a quadratic equation. It is derived from the general form \( ax^2 + bx + c = 0 \) and expressed as \( D = b^2 - 4ac \). For the given problem, the discriminant helps find conditions under which \( t_1 \) lies on the normal line. The equation obtained from the normal condition was \( tt_1^2 - 2t_1 + (t^3 - 2t) = 0 \).

• If \( D > 0 \), the quadratic equation has two distinct real roots.
• If \( D = 0 \), there is exactly one real root (roots are repeated).
• If \( D < 0 \), the equation has complex roots, indicating no intersection with real axes.

Finding susceptibility for real roots helps in determining the valid range for parameters, ensuring the solution is feasible in real-world scenarios.

Parametric Equations

Parametric equations allow us to express coordinates of points on a curve using parameters, making complex curves simpler to deal with. Instead of describing a parabola using Cartesian coordinates directly, we use parameters \( t \) and \( t_1 \) in our problem.

• The parametric representation \( P(t) = (t^2, 2t) \) and \( Q(t_1) = (t_1^2, 2t_1) \) translates complex algebra into more manageable equations.
• These equations link the x and y coordinates through parameters, easing curve-related calculations.
• Parametrics are not limited to parabolas; they are widely used in computer graphics, kinematics, and anywhere paths or trajectories are analyzed.

With parametric equations, transitions from one part of a curve to another are smooth, useful in handling dynamic systems and animations.