the tangents would be of the form $$y=m_1x-am_1^3-2am_1$$ and $$y=m_2x-am_2^3-2am_2$$ where $m_1$ and $m_2$ are the slopes of the respective tangents.
I think that this is not correct.
Let $(s,t)$ be a point on the parabola. Then, since $m=\frac{2a}{t}\Rightarrow t=\frac{2a}{m}$ with $t^2=4as$, $$y-t=\frac{2a}{t}(x-s)\iff y-\frac{2a}{m}=m\left(x-\frac{1}{4a}\left(\frac{2a}{m}\right)^2\right)$$So, $$y=m_1x+\frac{a}{m_1},\quad y=m_2x+\frac{a}{m_2}$$
I also know that $\tan 45=|\dfrac{m_1-m_2}{1-m_1m_2}|$
Note that we have $$\tan(45^\circ)=\left|\frac{m_1-m_2}{1\color{red}{+}m_1m_2}\right|\tag1$$
Since the intersection point $(X,Y)$ is $\left(\frac{a}{m_1m_2},\frac{m_1+m_2}{m_1m_2}a\right)$, we have $$m_1m_2=\frac{a}{X},\quad m_1+m_2=\frac YX.$$
Hence, from $(1)$ $$(1+m_1m_2)^2=(m_1-m_2)^2=(m_1+m_2)^2-4m_1m_2$$ $$\left(1+\frac aX\right)^2=\left(\frac YX\right)^2-4\cdot\frac aX,$$ i.e. $$Y^2-4aX=(X+a)^2$$
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