The tangents intercept a distance of $4c$ on the tangent at the vertex.
The third tangent at the vertex is the Y axis.
The point interception of tangents are $(0,4c)$ and $(0,-4c)$
Let them interestect at (h,k)
The equation of tangent to the parabola $$y=mx+\frac am$$ $$\pm 4c=0+\frac am$$ $$m=\frac {\pm a}{4c}$$
The slope of the first tangent is $$\frac{4c-k}{0-h}=\frac {a}{4c}$$ $$16c^2-4ck=-ah$$ $$ax-4cy+16c^2=0$$
But the answer given is $y^2-4ax=16c^2$
I know I have considered the other equation yet. I have it with me, but I know how to apply it.
Let the point of intersection be $P(h,k)$
which will satisfy $$y=mx+\dfrac am$$ i.e. $$k=mh+\dfrac am\implies m^2h-mk+a=0$$
if the two roots are $m_1,m_2$
$$m_1+m_2=\dfrac kh,m_1m_2=\dfrac ah\ \ \ \ (1)$$
Now the equation of the tangent through vertex $(0,0)$ is $x=0$ So, $y_k=\dfrac a{m_k}, k=1,2$
$$|4c|=|y_1-y_2|$$
$$\implies(4c)^2=a^2\dfrac{(m_1+m_2)^2-4m_1m_2}{(m_1m_2)^2}$$
Use $(1)$ to eliminate $m_1,m_2$