Final answer:
To find the equations of the tangents to the curve x = 9t^2 + 9, y = 6t^3 + 6 that pass through the point (18, 12), we first find the derivative of both x and y with respect to t. We use these derivatives to find the slopes of the tangents at t. Then, using the point-slope form of a line, we can write the equations of the tangents.
Step-by-step explanation:
To find the equations of the tangents to the curve x = 9t^2 + 9, y = 6t^3 + 6 that pass through the point (18, 12), we first find the derivative of both x and y with respect to t. We use these derivatives to find the slopes of the tangents at t. Then, using the point-slope form of a line, we can write the equations of the tangents.
The derivative of x with respect to t is dx/dt = 18t, and the derivative of y with respect to t is dy/dt = 18t^2. Since the slopes of the tangents at t are equal to the slopes of the tangent lines passing through (18, 12), we can set up the following equations:
dx/dt = 18t = (12 - y)/(18 - x)
dy/dt = 18t^2 = (y - 12)/(x - 18)
Solving these equations will give us the values of t at which the tangents pass through (18, 12). We substitute these values of t back into the original equations to find the corresponding x and y coordinates. Finally, using the point-slope form of a line, we can write the equations of the tangents.