91Ó°ÊÓ

To find the equation of a tangent line to a curve at a specific point, we work through a straightforward process. That point is where the line just touches the curve but doesn't cross it, thus being "tangent" to the curve. Here's how you can derive this equation easily: Begin with finding the slope of the tangent line. Once this part is done, use the point-slope form of the equation for a line: \[y - y_1 = m(x - x_1)\] In this formula, \(m\) stands for the slope you calculated, and \((x_1, y_1)\) is the point on the curve where the tangent line touches. For example, once we determine that at \((-1, 1)\) the slope \(m\) is \(-\frac{2}{3}\), the equation becomes \[y - 1 = -\frac{2}{3}(x + 1)\] This expression clearly represents the tangent line. To express it in a more standard linear equation form, like \(Ax + By + C = 0\), we can rearrange it to become \(3y + 2x - 5 = 0\). This way of presenting makes it easier to see the relationship between the line and the axes.

The concept of the slope is crucial in the context of tangent lines. The slope dictates how steep the line is, which helps us visualize how the tangent relates to the curve at a given point. For curves representing equations like \[x^4 - x^2y + y^4 = 1\], we use implicit differentiation to find the derivative, which gives us the slope of the line tangent to the curve. This process involves taking the entire expression and differentiating each term according to the rules of calculus, keeping in mind that both \(x\) and \(y\) are variables. Here, we set up the derivative expression and solve for \(\frac{dy}{dx}\) at the target point by substituting the coordinates. For example, substituting \((-1, 1)\), we get our slope \(m = -\frac{2}{3}\). Visualizing this, a negative slope like this suggests the line slants downwards from left to right as it touches the curve.

Verifying a point on a curve is a foundational step when working with tangent lines. This process involves checking if the point we are concerned with lies on the curve as defined by its equation. For our specific equation \[x^4 - x^2y + y^4 = 1\], we verify by plugging in the coordinates \((-1, 1)\) directly into the equation and ensure the left side equals the right side. If it checks out, the point is indeed on the curve. For this equation:

• \((-1)^4 = 1\)
• \((-1)^2 \times 1 = 1\)
• \(1^4 = 1\)

Substitute these results: \[1 - 1 + 1 = 1\]. Since both sides equal, the point \((-1, 1)\) is confirmed to be on the curve. This assurance allows us to accurately proceed with more complex tasks like finding tangent equations or calculating precise slopes related to that point.