The curve C has parametric equations x = t^3 – 8t

The curve C has parametric equations x = t^3 – 8t and y = t^2 where t is a parameter. Given that the point A has parameter t = –1, (a) find the coordinates of A. The line l is the tangent to C at A. (b) Show that an equation for l is 2x – 5y – 9 = 0.

The parametric equations of the curve C are x = t^3 – 8t and y = t^2, where t is a parameter. To find the coordinates of point A when t = -1, we substitute t = -1 into the equations. Therefore, x = (-1)^3 – 8(-1) = -1 + 8 = 7 and y = (-1)^2 = 1. Hence, the coordinates of point A are (7, 1).

The line l is the tangent to curve C at point A. To find the equation of the tangent line, we first need to find the derivative of y with respect to x. The derivative dy/dx = (dy/dt) / (dx/dt) = 2t / (3t^2 – 8). When t = -1, dy/dx = 2 / (-3) = -2/3. Therefore, the slope of the tangent line at point A is -2/3. Using the point-slope form of a linear equation, y – y1 = m(x – x1), where m is the slope and (x1, y1) is a point on the line, we substitute the values of A into the equation to get y – 1 = (-2/3)(x – 7). Simplifying this equation gives us 2x – 5y – 9 = 0, which is the equation of the tangent line l.

Therefore, the equation of the tangent line l to curve C at point A with coordinates (7, 1) is 2x – 5y – 9 = 0. This equation represents a straight line that touches the curve C at point A with a slope of -2/3. The tangent line l is perpendicular to the curve C at point A, and it intersects the curve at that point. This equation can be used to determine the relationship between the curve C and the tangent line l at point A.

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