a) The tangent to [tex]y=e^{ax}[/tex] at [tex]x=p[/tex] has slope
[tex]y' = ae^{ax} \implies y'(p) = ae^{ap}[/tex]
and [tex]y=e^{ap}[/tex] at this point. It passes through the origin, so its equation is
[tex]y - 0 = ae^{ap} (x - 0) \implies y = ae^{ap} x[/tex]
It also passes through the point [tex](p,e^{ap})[/tex] on the curve, so
[tex]y - e^{ap} = ae^{ap} (x - p) \implies y = e^{ap} + ae^{ap} x - ape^{ap}[/tex]
By substitution,
[tex]ae^{ap} x = e^{ap} + ae^{ap} x - ape^{ap} \implies e^{ap} = ape^{ap} \implies ap=1 \\\\ \implies \boxed{p=\dfrac1a}[/tex]
b) The normal to [tex]y=e^{ax}[/tex] at [tex]x=2p[/tex] has slope
[tex]-\dfrac1{y'(2p)} = -\dfrac1{ae^{2ap}} = -1 \implies ae^{2ap} = 1[/tex]
It follows that
[tex]ae^{2ap} = ae^2 = 1 \implies \boxed{a = \dfrac1{e^2} \text{ and } p = e^2}[/tex]
c) The tangent line equation is then
[tex]y = \dfrac1{e^2} e^1 x \implies \boxed{y = \dfrac xe}[/tex]
