I tried the general method of comparing separate equations ie. $$y=mx+\frac{1}{4m}$$ and $$y-3=mx\pm \sqrt 5( \sqrt {1+m^2})$$
Then $$\frac {1}{4m} =3 \pm \sqrt {5} (\sqrt {1+m^2})$$
I got stuck in the infinite operation of squaring the both sides.
However, even if I do eventually find the value of m after an eternity, it’s still an incredibly ineffecient way to solve. Is there a way to solve it more effectively, or is there an easier if squaring operation?
Hint:
The equation of any tangent at $(t^2,t)$
$$\dfrac{y-t}{x-t^2}=\dfrac1{2t}\iff x-2ty+t^2=0$$
Now if this has to be a tangent of the circle as well
the distance from the center has to be $=$radius
$$\implies\sqrt5=\dfrac{|0-2t\cdot3+t^2|}{\sqrt{1+(-2t)^2}}$$