Question 8: Because y'=2x, the tangent equation of passing point (AK, AK 2) is Y-AK 2 = 2ak (x-AK), and because the intersection of tangent and X axis is (a(k+ 1), 0), so A (k+ 1).

Question 10: Let P(x, y) be obtained by eliminating y from y=6cosx y=5tanx, and 6cosx = 5tanx =>6 (cosx) 2 = 5sinx.

=>6 (sinx) 2+5 sinx-6 = 0, = & gtSinx=-3/2 (truncated) or 2/3 because PP 1 is perpendicular to the X axis, and the point P, P 1, P2*** line.

So P 1P2=sinx=2/3.

Question 12: put 4

Multiply ① by ②, 2

hope this helps ...