Problem 143 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The normal at the point \(\left(b t_{1}^{2}, 2 b t_{1}\right)\) on a parabola meets the parabola again in the point \(\left(b t_{2}^{2}, 2 b t_{2}\right)\), then (a) \(t_{2}=t_{1}+\frac{2}{t_{1}}\) (b) \(t_{2}=-t_{1}-\frac{2}{t_{1}}\) (c) \(t_{2}=-t_{1}+\frac{2}{t_{1}}\) (d) \(t_{2}=t_{1}-\frac{2}{t_{1}}\)

Step-by-Step Solution

Answer

The correct answer is (c) \( t_2 = -t_1 + \frac{2}{t_1} \).

Step 1: Understand the geometry of the parabola

The standard parametric equation for a parabola is given as \( (bt^2, 2bt) \). Here, \( b \) is a constant and \( t \) is the parameter providing a point on the parabola.

Step 2: Find the slope of the tangent

For the point \( \left( b t_1^2, 2bt_1 \right) \) on the parabola, calculate the derivative to find the slope of the tangent: \( \frac{d(y)}{d(x)} = \frac{d(2bt)}{d(bt^2)} = \frac{2}{2t} = \frac{1}{t} \). At \( t = t_1 \), the slope of the tangent is \( \frac{1}{t_1} \).

Step 3: Determine the slope of the normal line

The slope of the normal is the negative reciprocal of the tangent's slope: \( -t_1 \). Thus, the equation of the normal at \( \left( b t_1^2, 2bt_1 \right) \) is \( y - 2bt_1 = -t_1(x - bt_1^2) \).

Step 4: Equation of the normal line

Simplify the normal equation: \( y - 2bt_1 = -t_1x + t_1^2b \). Therefore, the equation is \( y = -t_1x + t_1^2b + 2bt_1 \).

Step 5: Find intersection with the parabola

The normal equation meets the parabola again at \( \left( b t_2^2, 2bt_2 \right) \). Substitute these values into the normal equation:\( 2bt_2 = -t_1(b t_2^2) + t_1^2b + 2bt_1 \).

Step 6: Simplify and solve the quadratic equation

Simplify to obtain a relationship between \( t_1 \) and \( t_2 \). Rearrange terms in \(-bt_1t_2^2 + 2bt_2 - t_1^2b - 2bt_1 = 0 \) and solve the quadratic in \( t_2 \):\( t_1t_2^2 - 2t_2 + t_1 = 0 \).Now, using Vieta's formulas for roots, the sum of roots \( t_1 + t_2 = \frac{2}{t_1} \), so, solving for \( t_2 \) gives: \( t_2 = -t_1 + \frac{2}{t_1} \).

Step 7: Compare with given choices

Comparing our result \( t_2 = -t_1 + \frac{2}{t_1} \) with the options provided, the correct answer is option (c).

Key Concepts

Normal to Parabola

A parabola is a curve where any point is equidistant from a fixed point called the focus and a line called the directrix. The **normal** to a parabola at a given point is a line perpendicular to the tangent line at that point. To find the equation of the normal line to a parabola at a point, you first need to calculate the slope of the tangent. If a parabola is given in the parametric form as y = f(x), the normal line can be determined by first finding the derivative, which gives the slope of the tangent: \( ext{slope of tangent} = \frac{dy}{dx} \). The slope of the normal is the negative reciprocal of the slope of the tangent. Consider a point on the parabola parameterized as \( (bt^2, 2bt)\). To find the slope of the normal at this point, \( ext{slope of normal} = -\frac{1}{\text{slope of tangent}}= -t \). Thus, using the point-slope form of a line, the equation of the normal can be expressed as: \( y - y_1 = -tx + x_1\)(where \( (x_1, y_1) \) is the point on the parabola). This concept is crucial in finding how the normal intersects the parabola in other points.

Tangent to Parabola

The **tangent** to a parabola is a straight line that touches the curve at exactly one point without crossing it. The slope of this line at any given point on the parabola can be calculated using the derivative of the parabola's equation. For parabolas given in parametric form, such as \( (bt^2, 2bt)\), taking the derivative yields the slope of the tangent line. In this specific exercise, it works out to be \( \frac{1}{t} \). Thus, for a point on the parabola at \( t = t_1 \), the slope of the tangent is \( \frac{1}{t_1} \).The tangent line's equation is usually written in point-slope form: \( y - y_1 = m(x - x_1) \). Using the calculated slope and the given point, this equation becomes: \( y - 2bt_1 = \frac{1}{t_1}(x - bt_1^2) \). Understanding the tangent's equation is vital as it forms the basis for identifying how the normal line, coming from the same parabola point, intersects again with the parabola.

Parametric Equations

**Parametric equations** are used to express the coordinates of the points that make up a geometric object, such as a curve, in terms of a parameter. In the context of a parabola, these equations allow you to explore points on the parabola by varying the parameter. For the parabola in question, given in the form \( (bt^2, 2bt) \), \( t \) is the parameter that helps describe every point on the parabola as \( x = bt^2 \) and \( y = 2bt \). Parametric equations are incredibly helpful because they provide a straightforward way to derive various calculus-related properties such as tangents and normals.When solving for intersections or analyzing the behavior of curves, parametric forms simplify the process. They allow modifications of one variable while assessing its impact on both the x-y coordinates simultaneously. Understanding parametrics is not only useful for comprehending the exercise at hand, where the intersection of a normal with the same parabola is determined, but also for a wide array of other applications in physics and engineering.