Question

The tangent of the graph of y=
`e^(ax)`, a
`\\eq`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=\frac1a}`

b) The normal to
`y=e^(ax)` at
`x=2p` has slope

`-\frac1{y'(2p)} = -\frac1{ae^(2ap)} = -1 \implies ae^(2ap) = 1`

It follows that

`ae^(2ap) = ae^2 = 1 \implies \boxed{a = \frac1{e^2} \text{ and } p = e^2}`

c) The tangent line equation is then

`y = \frac1{e^2} e^1 x \implies \boxed{y = \frac xe}`