Final answer:
The points on curve C where the tangent line has a slope of 1 are found by implicitly differentiating the equation of the curve and setting the derivative equal to 1. The resulting points on the curve are (0, ±1/√2).
Step-by-step explanation:
The student is asked to find all points (x,y) on the curve C implicitly defined by (x² - 4xy + 2y² = 1) where the tangent line has a slope of 1. To solve this, we need to use implicit differentiation to find dy/dx, which represents the slope of the tangent line. Once we have the derivative, we set it equal to 1 to satisfy the condition that the slope must be 1 and solve for the points (x,y).
We begin by differentiating both sides of the given equation with respect to x:
- For the x² term, we get 2x.
- For the -4xy term, applying the product rule, we get -4y - 4x(dy/dx).
- For the 2y² term, differentiating implicitly with respect to x, gives us 4y(dy/dx).
After differentiating, we set the derivative equal to 1 and solve for yrevelations to find the specific points:
- 2x - 4y - 4x(dy/dx) + 4y(dy/dx) = 0
- - 4y + 4y(dy/dx) = -2x (Rearranging and factoring out common terms)
- dy/dx = (4y - 2x)/(4y) (After simplifying)
- 1 = (4y - 2x)/(4y) (Since we are looking for a slope of 1)
- 4y = 4y - 2x (Multiplying both sides by 4y)
- x = 0 (Solving for x yields x = 0 since all y terms cancel out)
Finally, substituting x = 0 back into the original equation yields y = ±1/√2. Thus, the points on the curve C where the tangent line has a slope of 1 are (0, ±1/√2).