x = 2/t
y = t² - 5t + 4

Where the curve cuts the x-axis,
y = 0
t² - 5t + 4 = 0
(t - 1) (t - 4) = 0
t = 1 or t = 4

You will need these for the two different tangents possible.

Now, look at the original equation again.

dx/dt = -2/t²
dy/dt = 2t - 5

Case 1: t = 1

Here, you will have

x = 2/1 = 2
y = 0 (obviously; that's how you got t = 1 in the first place)

dx/dt = -2
dy/dt = -3
dy/dx = -3/-2 = 3/2

So, the tangent is a line of gradient 3/2 passing through (2, 0).

y = 1.5x + c
0 = 1.5 (2) + c
0 = 3 + c
-3 = c
y = 1.5x - 3

Case 2: t = 4

Here, you will have

x = 2/4 = 1/2
y = 0 (obviously; that's how you got t = 4 in the first place)

dx/dt = -1/8
dy/dt = 3
dy/dx = 3/(-1/8) = -24

So, the tangent is a line of gradient -24 passing through (1/2, 0).

y = -24x + c
0 = -24 (1/2) + c
0 = -12 + c
12 = c
y = -24x + 12