Problem 102 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

The tangents to the curve \(y=(x-2)^{2}-1\) at its points of intersection with the line \(x-y=3\), intersect at the point : (b) \(\left(-\frac{5}{2},-1\right)\) (a) \(\left(\frac{5}{2}, 1\right)\) (c) \(\left(\frac{5}{2},-1\right)\) (d) \(\left(-\frac{5}{2}, 1\right)\)

Step-by-Step Solution

Answer

The tangents intersect at \(\left(\frac{5}{2}, -1\right)\), option (c).

Step 1: Find Intersection Points

First, we need to find the intersection points of the given curve and the line. The curve is \(y = (x-2)^2 - 1\) and the line is \(x - y = 3\). Substitute the expression for \(y\) from the curve equation into the line equation: \[ x - ((x-2)^2 - 1) = 3 \] Simplifying, we get \[ x - (x^2 - 4x + 4) + 1 = 3 \] which simplifies to \[ -x^2 + 5x - 5 = 3 \]. This further simplifies to \[-x^2 + 5x - 8 = 0\]. Multiply by -1: \[ x^2 - 5x + 8 = 0 \]. Factor or use the quadratic formula to find \(x\).

Step 2: Solve Quadratic

The quadratic equation from Step 1 is \(x^2 - 5x + 8 = 0\). Since this equation cannot be factored easily, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1, b = -5, c = 8\). Then \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} \] which simplifies to \[ x = \frac{5 \pm \sqrt{1}}{2} \]. So, \(x = 3\) and \(x = 2\).

Step 3: Find Corresponding y-values

Using the \(x\)-values found in Step 2, substitute back into the curve equation \(y = (x-2)^2 - 1\) to find the corresponding \(y\)-values for each \(x\). For \(x = 3\), \(y = (3-2)^2 - 1 = 0\). For \(x = 2\), \(y = (2-2)^2 - 1 = -1\). The intersection points are \((3, 0)\) and \((2, -1)\).

Step 4: Determine Tangent Lines

To find the equations of the tangents at the intersection points, first find the derivative of the curve \(y = (x-2)^2 - 1\). The derivative is \(y' = 2(x-2)\). For \(x = 3\), \(y' = 2(3-2) = 2\). For \(x = 2\), \(y' = 2(2-2) = 0\). Thus, the tangents are \(y - 0 = 2(x - 3)\) leading to \(y = 2x - 6\), and \(y + 1 = 0(x - 2)\) resulting in \(y = -1\).

Step 5: Solve Intersection of Tangents

Solve the system of equations formed by the tangents from Step 4 to find their intersection. The equations are: \(y = 2x - 6\) and \(y = -1\). Set them equal to each other: \[ 2x - 6 = -1 \] which simplifies to \[ 2x = 5 \], thus \(x = \frac{5}{2}\). Substitute \(x = \frac{5}{2}\) back into \(y = -1\) to confirm the \(y\)-coordinate. So the intersection of the tangents is \(\left(\frac{5}{2}, -1\right)\).

Key Concepts

Intersection Points

Understanding the concept of intersection points can significantly clarify how different mathematical objects relate to each other. When we talk about the intersection of a curve and a line, we mean the points at which the curve and the line meet in the coordinate plane.
In this exercise, the curve is given by the equation \( y = (x-2)^2 - 1 \) and the line is given by \( x-y = 3 \). To find the intersection points, we substitute the expression for \( y \) from the curve into the equation of the line, resulting in a single equation in terms of \( x \).
Solving this equation offers the \( x \)-coordinates where the curve and the line intersect. We next find the corresponding \( y \)-values by substituting these \( x \)-values back into either the curve equation or the line equation. This process provides the full coordinates of the intersection points, which are vital for further analysis in calculating tangents.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It is particularly useful when an equation cannot be easily factored.
The formula is given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation. In the context of our exercise, the quadratic arises from substituting the curve equation into the line equation, resulting in \( x^2 - 5x + 8 = 0 \).
Applying the quadratic formula to this equation, we calculate the roots as \( x = 3 \) and \( x = 2 \). These roots are critical because they give us the \( x \)-coordinates of the intersection points.

Derivative of a Function

The derivative of a function represents the rate of change or the slope of the function's graph at a particular point. It is a cornerstone concept in calculus.
For the function given in our exercise \( y = (x-2)^2 - 1 \), we calculate the derivative to figure out the slopes of tangent lines at specific points. The derivative is found using the power rule, resulting in \( y' = 2(x-2) \).
By evaluating this derivative at specific \( x \)-coordinates (from our intersection points), we obtain the slopes needed to form the lines tangent to the curve at those points. Thus, these derivatives provide guidance on how the curve behaves locally at the intersection points.

Equation of a Tangent Line

The equation of a tangent line tells us how a line just 'touches' a curve at a given point. This line has the same slope as the curve at that point and represents the best local linear approximation of the curve.
To find this equation, we use the point-slope form of a linear equation \( y - y_1 = m(x - x_1) \), where \( (x_1, y_1) \) is the point of tangency, and \( m \) is the slope given by the derivative of the curve at \( x_1 \).
In our problem, the tangents at the intersection points \( (3, 0) \) and \( (2, -1) \) are derived by substituting these points and their respective slopes (calculated using the derivative) into this form. This results in equations \( y = 2x - 6 \) and \( y = -1 \), both representing lines that just "kiss" the curve at the interactive set points.