Finding Point P, Tangent, And Normal Equations On A Curve

This article delves into a fascinating problem in coordinate geometry, exploring the intersection of lines and curves. Specifically, we aim to determine the coordinates of a point P where a line perpendicular to a given line intersects a curve. Additionally, we will derive the equations of the tangent and normal lines to the curve at the points of intersection. This exploration will enhance your understanding of geometric relationships, line equations, curve analysis, and calculus applications in geometry. Understanding these concepts is crucial for students and enthusiasts alike, as they form the backbone of many advanced mathematical and engineering applications. By meticulously dissecting this problem, we will not only find the solution but also illuminate the underlying principles that make coordinate geometry a powerful tool for problem-solving.

Problem Statement

The problem at hand involves several key elements: a line, a curve, and a perpendicular relationship. We are given a line that passes through the point $(1, 6)$ and is perpendicular to the line $x - y = 5$. This line intersects the curve defined by the equation $y = 2x + 4x^{-1}$ at two points. Our primary objective is to find the coordinates of the second point of intersection, which we denote as P. Once we locate the coordinates of point P, we will then proceed to determine the equations of the tangent and normal lines to the curve at both points of intersection. This comprehensive analysis will not only provide a concrete solution but also illustrate the interplay between algebraic equations and geometric properties. Each step in the solution process will be carefully explained, ensuring that the underlying logic and methodologies are clear and accessible.

1. Determining the Equation of the Perpendicular Line

To begin, we need to find the equation of the line that passes through the point $(1, 6)$ and is perpendicular to the line $x - y = 5$. Let's break this down step by step to ensure clarity and accuracy. First, we rewrite the given equation $x - y = 5$ in the slope-intercept form, which is $y = mx + c$, where m represents the slope and c is the y-intercept. Rearranging the equation, we get $y = x - 5$. From this form, it is evident that the slope of the given line is $m_1 = 1$. Since we are looking for a line perpendicular to this, we need to find the slope of the perpendicular line. The slopes of two perpendicular lines are negative reciprocals of each other. Therefore, if the slope of the given line is $m_1$, the slope of the perpendicular line, $m_2$, is given by m_2 = -rac{1}{m_1}. Substituting $m_1 = 1$, we find that $m_2 = -1$. Now that we have the slope of the perpendicular line, we can use the point-slope form of a line equation, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and m is the slope. We are given the point $(1, 6)$ through which the perpendicular line passes. Plugging in the values, we get $y - 6 = -1(x - 1)$. Simplifying this equation, we have $y - 6 = -x + 1$. Further rearranging to get the equation in slope-intercept form, we have $y = -x + 7$. Thus, the equation of the line perpendicular to $x - y = 5$ and passing through $(1, 6)$ is $y = -x + 7$. This equation is crucial as it sets the stage for finding the points of intersection with the given curve. Understanding how to derive this equation is fundamental for solving various problems in coordinate geometry.

2. Finding the Points of Intersection

Having established the equation of the perpendicular line as $y = -x + 7$, our next crucial step is to determine the points where this line intersects the curve defined by the equation $y = 2x + 4x^{-1}$. To achieve this, we need to solve these two equations simultaneously. This involves setting the expressions for y from both equations equal to each other, effectively finding the x-values where the line and the curve meet. Thus, we equate $-x + 7$ to $2x + 4x^{-1}$, giving us the equation: $-x + 7 = 2x + 4x^{-1}$. This equation now represents the condition for the intersection points. To solve this equation, we first eliminate the fraction by multiplying the entire equation by x. This gives us a new equation: $x(-x + 7) = x(2x + 4x^{-1})$, which simplifies to $-x^2 + 7x = 2x^2 + 4$. Now, we rearrange the terms to form a quadratic equation, which is a standard form for solving such problems. Moving all terms to one side, we get $0 = 2x^2 + x^2 - 7x + 4$, which simplifies to the quadratic equation $3x^2 - 7x + 4 = 0$. To find the solutions for x, we can either factorize the quadratic equation or use the quadratic formula. In this case, the equation can be factored as $(3x - 4)(x - 1) = 0$. This factorization leads us to two possible values for x: $x = 1$ and x = rac{4}{3}. These x-values represent the x-coordinates of the points where the line and the curve intersect. To find the corresponding y-coordinates, we substitute these x-values back into either the equation of the line or the equation of the curve. It is often simpler to use the linear equation $y = -x + 7$. For $x = 1$, we have $y = -1 + 7 = 6$. This gives us the point $(1, 6)$, which we already knew was a point of intersection. For x = rac{4}{3}, we have y = -rac{4}{3} + 7 = rac{-4 + 21}{3} = rac{17}{3}. Thus, the second point of intersection, which we denoted as P, has coordinates $\left(\frac{4}{3}, \frac{17}{3}\right)$.

3. Determining the Tangent and Normal Equations

With the coordinates of the intersection points established, our next objective is to find the equations of the tangent and normal lines to the curve $y = 2x + 4x^{-1}$ at these points. This involves using differential calculus to determine the slope of the curve at a given point, which will then allow us to construct the equations of the tangent and normal lines. The first step in this process is to find the derivative of the curve's equation, which represents the slope of the curve at any point. The given curve is $y = 2x + 4x^{-1}$. Differentiating y with respect to x, we use the power rule of differentiation. The derivative of $2x$ is simply 2. For the term $4x^{-1}$, we apply the power rule, which states that the derivative of $x^n$ is $nx^{n-1}$. Thus, the derivative of $4x^{-1}$ is $4(-1)x^{-2} = -4x^{-2}$. Combining these, we get the derivative $\frac{dy}{dx} = 2 - 4x^{-2}$, which can also be written as $\frac{dy}{dx} = 2 - \frac{4}{x^2}$. This derivative gives us the slope of the tangent line at any point on the curve. Now, we need to find the slopes at the specific points of interest, which are $(1, 6)$ and $\left(\frac{4}{3}, \frac{17}{3}\right)$. At the point $(1, 6)$, we substitute $x = 1$ into the derivative: $\frac{dy}{dx}|_{x=1} = 2 - \frac{4}{1^2} = 2 - 4 = -2$. This means the slope of the tangent line at $(1, 6)$ is -2. The slope of the normal line, being perpendicular to the tangent, will be the negative reciprocal of the tangent's slope. Thus, the slope of the normal at $(1, 6)$ is $\frac{1}{2}$. At the point $\left(\frac{4}{3}, \frac{17}{3}\right)$, we substitute $x = \frac{4}{3}$ into the derivative: $\frac{dy}{dx}|_{x=\frac{4}{3}} = 2 - \frac{4}{\left(\frac{4}{3}\right)^2} = 2 - \frac{4}{\frac{16}{9}} = 2 - \frac{4 \times 9}{16} = 2 - \frac{9}{4} = \frac{8 - 9}{4} = -\frac{1}{4}$. So, the slope of the tangent line at $\left(\frac{4}{3}, \frac{17}{3}\right)$ is $-\frac{1}{4}$. The slope of the normal line at this point is the negative reciprocal, which is 4. Now that we have the slopes of the tangents and normals at both points, we can use the point-slope form of a line equation to find their equations. The point-slope form is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and m is the slope. For the tangent at $(1, 6)$ with slope -2, the equation is $y - 6 = -2(x - 1)$, which simplifies to $y = -2x + 8$. For the normal at $(1, 6)$ with slope $\frac{1}{2}$, the equation is $y - 6 = \frac{1}{2}(x - 1)$, which simplifies to $y = \frac{1}{2}x + \frac{11}{2}$. For the tangent at $\left(\frac{4}{3}, \frac{17}{3}\right)$ with slope $-\frac{1}{4}$, the equation is $y - \frac{17}{3} = -\frac{1}{4}\left(x - \frac{4}{3}\right)$, which simplifies to $y = -\frac{1}{4}x + 6$. For the normal at $\left(\frac{4}{3}, \frac{17}{3}\right)$ with slope 4, the equation is $y - \frac{17}{3} = 4\left(x - \frac{4}{3}\right)$, which simplifies to $y = 4x + \frac{1}{3}$. Thus, we have found the equations of the tangent and normal lines at both points of intersection.

4. Summary of Results

To provide a clear overview of our findings, let's summarize the results of our analysis. We started with the problem of finding the coordinates of the point P where a line perpendicular to $x - y = 5$ and passing through $(1, 6)$ intersects the curve $y = 2x + 4x^{-1}$. We also aimed to determine the equations of the tangent and normal lines to the curve at the intersection points. Through a series of steps involving coordinate geometry and differential calculus, we have successfully achieved these objectives.

Coordinates of Point P:

We found that the line perpendicular to $x - y = 5$ and passing through $(1, 6)$ has the equation $y = -x + 7$. By solving this equation simultaneously with the curve equation $y = 2x + 4x^{-1}$, we identified the points of intersection. One point was the given $(1, 6)$, and the other point, P, has coordinates $\left(\frac{4}{3}, \frac{17}{3}\right)$.

Tangent and Normal Equations at (1, 6):

At the point $(1, 6)$, we calculated the slope of the tangent line to be -2. Using the point-slope form, we found the equation of the tangent line to be $y = -2x + 8$. The slope of the normal line, being perpendicular to the tangent, is $\frac{1}{2}$. Thus, the equation of the normal line at $(1, 6)$ is $y = \frac{1}{2}x + \frac{11}{2}$.

Tangent and Normal Equations at P ($\frac{4}{3}$, $\frac{17}{3}$):

At the point P, which has coordinates $\left(\frac{4}{3}, \frac{17}{3}\right)$, we calculated the slope of the tangent line to be $-\frac{1}{4}$. The equation of the tangent line at this point is $y = -\frac{1}{4}x + 6$. The slope of the normal line at P is 4, making the equation of the normal line $y = 4x + \frac{1}{3}$. This summary encapsulates the main results of our problem-solving process, providing a clear and concise overview of the coordinates of point P and the equations of the tangent and normal lines at both intersection points. The combination of algebraic manipulation and calculus techniques has allowed us to thoroughly analyze the geometric relationships in this problem. These results underscore the interconnectedness of different mathematical concepts in solving complex problems.

Conclusion

In conclusion, this exploration has successfully demonstrated the process of finding intersection points between a line and a curve, as well as determining the equations of tangent and normal lines at those points. By applying principles of coordinate geometry and differential calculus, we were able to systematically solve the problem. The determination of the line's equation, the algebraic solution for intersection points, and the calculation of tangent and normal lines showcase the power and elegance of mathematical problem-solving. The skills and concepts utilized in this analysis are fundamental in various fields, including engineering, physics, and computer graphics, where understanding geometric relationships and curve behavior is crucial. The ability to manipulate equations, find derivatives, and interpret geometric meanings from algebraic expressions is a valuable asset for any aspiring mathematician or scientist. Through meticulous steps and clear explanations, this article has not only provided a solution to the specific problem but also illuminated the broader principles that govern mathematical analysis and problem-solving strategies. The reader can now apply these techniques to similar problems, reinforcing their understanding and enhancing their mathematical proficiency. This exercise serves as a testament to the importance of combining theoretical knowledge with practical application, fostering a deeper appreciation for the beauty and utility of mathematics.