i)
x = t² - 2t, y = ln t
dx/dt = 2t - 2 , dy/dt = 1/t , t > 0

dy/dx = dy/dt × dt/dx
= dy/dt ÷ dx/dt
= 1/(t(2t -2)
= 1/(2t² - 2t)

dy/dx is undefined when tangent is parallel to y axis. This means the denominator is 0.

So 2t² - 2t = 0
t(t -1) = 0
t = 0 (N.A as t > 0) or t = 1


ii)

when t = 3, gradient of tangent
= 1/(2(3²) - 2(3))
= 1/12

x = 3² - 2(3) = 3
y = ln 3

Sub into equation y = mx + c,
ln 3 = (1/12)(3) + c
c = ln3 - ¼
So equation of tangent : y = 1/12 x + ln3 - ¼


Gradient of normal = -1/(1/12) = -12
Sub into equation y = mx + c
ln 3 = -12(3) + c
c = ln3 + 36
So equation of tangent : y = -12x + ln3 + 36


The triangle bound by the y-axis and the two lines has :

① A base which is part of the y axis with a length equal to the distance between the two y intercepts of the two lines.

② a perpendicular height which is equal to the horizontal distance between the y-axis and the point (3,ln3). i.e 3 units, distance from x = 0 to x = 3)


So area of triangle

= ½ x base x height
= ½ x ( ln3 + 36 - (ln3 - ¼) )units x 3 units
= ½ x 145/4 units x 3 units
= 435/8 units²

(Shown)