How to find the nearest and farthest point of a conic with a given point P

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Given the following equation : $4x^2 - 9y^2 = 1$ And the point $P=(2,1)$ I have to find the nearest and farthest point of the conic with the point $P$.

What I have done:

I set $x = 1/2\cosh(t)$ and $y = 1/3\sinh(t)$

Then I calculated the squared distance $d^2 = (x-2)^2 + (y-1)^2$ so I have the equation $f(t) = ....$ by replacing $x$ and $y$.

I got $$f(t) = {13\over72}e^{2t} - {13\over72}e^{-2t} -{4\over3}e^t +{2\over3}e^{-t}$$ because I replaced cosh(t) as (e^t+e^-t)/2 and sinh(t) as (e^t-e^-t)/2

Then I have to find the derivate $f'(t)$ and and equate it to $0$, and I can find the maximum and minimum (if exists) of f(t.) I got $$f'(t) = {26\over72}e^{2t} + {26\over72}e^{-2t} -{4\over3}e^t -{2\over3}e^{-t}$$ I'm stuck at solving $f'(t) = 0$.

Any help would be appreciated. Thank you.