Question

The parametric equations of a curve are x = at, y = at³ where a is a positive constant. The point P on the curve has parameter p, and the tangent to the curve at point P cuts the y−axis at S and the x−axis at T. The point M is the midpoint of ST. Find a Cartesian equation of the curve traced by M as p varies.

a) y = x²
b) y = x³
c) y = 2x³
d) y = 3x²

Answer 1

The Cartesian equation of the curve traced by midpoint M as p varies is not directly listed among the given options, but the process involves finding the derivatives, applying the slope of the tangent, determining points S and T, and then eliminating p to find the relationship between x and y for M's trajectory.

Step-by-step explanation:

We need to find the Cartesian equation of the curve that the midpoint M of ST traces as p varies. We start by finding the derivatives dx/dt and dy/dt to obtain the slope of the tangent at point P:

• dx/dt = a
• dy/dt = 3a(p)^2

Thus, the slope of the tangent at P is given by dy/dt divided by dx/dt, which is 3a(p)^2 / a = 3p^2. Using the point-slope formula, the equation of the tangent line at P (with coordinates (ap, a(p)^3)) is:

y - a(p)^3 = 3p^2 (x - ap)

This line cuts the y-axis where x = 0 (point S), and the x-axis where y = 0 (point T). Solving for the points S and T:

• When x = 0, y = a(p)^3 - 3a(p)^3 = -2a(p)^3 (point S)
• When y = 0, x = ap / (1 - 3p^2) (point T, assuming 3p^2 ≠ 1)

The midpoint M of ST has an x-coordinate (ap / (1 - 3p^2))/2 and a y-coordinate of -a(p)^3. Hence, the Cartesian equation for M, as p varies, can be found by eliminating p.

The x-coordinate can be rewritten in terms of p as x = ap / (1 - 3p^2). We rearrange it to p = x / (a - 3ax^2). Squaring p, we get p^2 = x^2 / (a^2 - 6a^2x^2 + 9a^2x^4). The y-coordinate of M is -a(p)^3, so substituting p^2 we get:

y = -ax^3 / ((a^2 - 6a^2x^2 + 9a^2x^4)^(3/2))

For this to be true for all values of a, it must be the case that the curve traced by M as p varies is y = -x^3. However, this does not match any of the given options.