I am given that $x=ae^{-t}$ and that $y=be^{2t}.$ I'm asked to find the tangent line at $t=0.$
I have said that $$\frac{dx}{dt}=-ae^{-t}, \frac{dy}{dt}=2be^{2t}$$
Thus $$\frac{dy}{dx}=\frac{-2b}{a}e^{3t}=\frac{-2b}{a} \text{ at } t=0$$
When $t=0, x=a, y=b.$
So we have that the tangent line is $$y-b=\frac{-2b}{a}(x-a)$$
However, the solution says that the answer should be $y=x-1.$
I really don't see how they've got this. Could someone explain?
Your work is correct and the solution key is wrong. Here is a graph of the parametric curve (blue), your tangent line (red, in simplified form), and the solution key's line (green). The cross-hairs is a trace at $t=0$. This example has $a=1,\ b=2$. Your line is clearly correct, the solution key's is clearly wrong.
More people should be using a grapher as a check! (This is from my TI-Npire CX Teacher Software.)