Question

The tangent of the graph of y=="absmiddle" class="latex-formula">, a$\neq$0, at a point x=p passes through the origin.
a) Express the value of p in terms of a
b) If the gradient of the normal to the curve at the point x=2p is -1, find the values of a and p.
c) Hence, find the equation of the tangent to the curve at the point x=p.

Answer 1

a) The tangent to $y=e^{ax}$ at $x=p$ has slope

$y' = ae^{ax} \implies y'(p) = ae^{ap}$

and $y=e^{ap}$ at this point. It passes through the origin, so its equation is

$y - 0 = ae^{ap} (x - 0) \implies y = ae^{ap} x$

It also passes through the point $(p,e^{ap})$ on the curve, so

$y - e^{ap} = ae^{ap} (x - p) \implies y = e^{ap} + ae^{ap} x - ape^{ap}$

By substitution,

$ae^{ap} x = e^{ap} + ae^{ap} x - ape^{ap} \implies e^{ap} = ape^{ap} \implies ap=1 \\\\ \implies \boxed{p=\dfrac1a}$

b) The normal to $y=e^{ax}$ at $x=2p$ has slope

$-\dfrac1{y'(2p)} = -\dfrac1{ae^{2ap}} = -1 \implies ae^{2ap} = 1$

It follows that

$ae^{2ap} = ae^2 = 1 \implies \boxed{a = \dfrac1{e^2} \text{ and } p = e^2}$

c) The tangent line equation is then

$y = \dfrac1{e^2} e^1 x \implies \boxed{y = \dfrac xe}$