The parametric equations of a curve are x = etant, y = 3 tan²t. Find the equation of the tangent to the curve at the point (e, 3). Give your answer in the form y = mx+c, where m and c are exact.

Mark scheme

Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.

1. [B1] At (e, 3), tan t = 1 or t = π/4 SOI
2. [B1] dx/dt = sec²t × e^tant
3. [B1] dy/dt = 6 tan t × sec²t
4. [*M1] dy/dx = (6 tan t × sec²t) / (sec²t × e^tan t) (= 6 tan t × e^-tan t) Correct use of their derivatives.
5. [DM1] Substitute for t, and use correct method for the equation of the line.
6. [A1] y = (6/e)x - 3 Or exact equivalent.
7. Alternative Method for Question 4
8. [B1] Correct cartesian form, e.g. x = e^sqrt(y/3) or ln x = sqrt(y/3) or y = 3(ln x)² SOI
9. [*M1] Differentiate function of a function Complete method.
10. [A1] Obtain dy/dx = k*(1/x)*(y/sqrt(3)) or dy/dx = 6x^-1 lnx
11. [DM1] Use correct method for the equation of the line.
12. [A1] y = (6/e)x - 3