Final answer:
To find the equation of the tangent to the curve at the point corresponding to t = 1, we need to find the slope of the curve at that point. By differentiating the equations x = t - t⁻¹ and y = 8t² with respect to t and evaluating the derivatives at t = 1, we find the slope is 2. Using the point-slope form of a line, we can then find the equation of the tangent.
Step-by-step explanation:
To find the equation of the tangent to the curve at the point corresponding to t = 1, we need to find the slope of the curve at that point. Given x = t - t⁻¹ and y = 8t², we can differentiate both equations with respect to t. The derivative of x is 1 - (-1/t²) = 1 + 1/t² and the derivative of y is 16t. Evaluating these derivatives at t = 1, we get the slope of the curve at that point: m = 1 + 1/1² = 2. Now, we can use the point-slope form of a line to find the equation of the tangent. Substituting the values of x, y, and m into the equation y - y₁ = m(x - x₁), we get y - 8 = 2(x - 1). Simplifying, we have y = 2x - 6 as the equation of the tangent.