91Ó°ÊÓ

Curve analysis is a fundamental concept in calculus and differential equations. It helps us understand the behavior and properties of curves represented by functions. In this particular problem, we analyze the properties of a curve where the point \(\text{P}\) lies on it, and we use the relationship between the tangent and normal lines at this point to derive a differential equation. Understanding the relative positions of points and lengths on the curve is crucial. Given \(\text{P}(x, y)\), we use the distance formula \(\text{OP} = \sqrt{x^2 + y^2}\) to find the length from the origin to the point. The analysis begins by setting up relationships and using geometrical properties, such as gradients and intersection points. Understanding curve behavior allows us to establish key relationships and form the differential equations needed to solve for the curve's equation. Taking each step methodically ensures accuracy and comprehension.

The concept of a normal to a curve is essential in this exercise. A normal line at any point \(\text{P}\) on a curve is perpendicular to the tangent at that point. Let the gradient of the curve at \(\text{P}\) be given by \(\frac{dy}{dx}\). Consequently, the gradient of the normal line will be the negative reciprocal of this gradient, denoted as \(-\frac{dx}{dy} = -\frac{1}{\frac{dy}{dx}}\). Knowing this, we can determine where the normal intersects the \(x\)-axis. For point \(\text{P}(x, y)\), the coordinates of the intersection point \(\text{Q}(x_1, 0)\) can be found using the normal gradient: \(x_1 = x - y\frac{dx}{dy}\). This allows us to further analyze the geometrical properties of the curve and establish the necessary equations. This specificity in understanding and using normal lines helps solve complex problems involving distances and relationships on curves.

Integration techniques are vital in solving differential equations, which allow us to find the explicit form of the curve. Once we have the differential equation, \( y \frac{dy}{dx} = x \), we rewrite it in a form suitable for integration: \ y dy = x dx \. This separated variables form is key. Integrating both sides involves finding the integrals: \ \ \int y \ dy = \ \int x \ dx \. These integrals are straightforward: \int y dy = \frac{y^2}{2}\ and \ \int x dx = \frac{x^2}{2} \. Solving yields the general solution \ \frac{y^2}{2} = \ \frac{x^2}{2} + C\ , which simplifies to \ y^2 = x^2 + C \. Understanding and applying these integration techniques allow for the transition from a differential equation representing a relationship to an explicit equation representing the curve's behavior. Mastery of these techniques is crucial for solving numerous problems in calculus and differential equations.