Answer
The path is ${\bf{r}}\left( t \right) = \left( {t,2{t^4}} \right)$.
Work Step by Step
We are given $f\left( {x,y} \right) = 2{x^2} + 8{y^2}$. The gradient is
$\nabla f = \left( {4x,16y} \right)$
Let a path of the form ${\bf{r}}\left( t \right) = \left( {t,g\left( t \right)} \right)$ passing through $\left( {1,2} \right)$ follow the gradient of $f\left( {x,y} \right) = 2{x^2} + 8{y^2}$. Then, the tangent vector ${\bf{r}}'\left( t \right)$ points in the direction of $\nabla f$ for all $t$. That is, ${\bf{r}}'\left( t \right) = k\left( t \right)\nabla {f_{{\bf{r}}\left( t \right)}}$ for some positive scalar function $k\left( t \right)$.
So, the equation ${\bf{r}}'\left( t \right) = k\left( t \right)\nabla {f_{{\bf{r}}\left( t \right)}}$ becomes
$\left( {1,g'\left( t \right)} \right) = k\left( t \right)\left( {4t,16g\left( t \right)} \right)$
In component forms, we get
$1 = 4tk\left( t \right)$,
$g'\left( t \right) = 16k\left( t \right)g\left( t \right)$
From the first equation we obtain $k\left( t \right) = \frac{1}{{4t}}$.
Substituting it in the second equation above gives
$g'\left( t \right) = \frac{4}{t}g\left( t \right)$
Since $g'\left( t \right) = \frac{{dg}}{{dt}}$, so
$\frac{{dg}}{g} = 4\frac{{dt}}{t}$
Integrating both sides gives
$\smallint \frac{{{\rm{d}}g}}{g} = 4\smallint \frac{{{\rm{d}}t}}{t}$
$\ln g = 4\ln t + \ln c$, ${\ \ \ }$ $\ln g = \ln \left( {c{t^4}} \right)$,
where $c$ is integration constant.
Thus, the solution is $g\left( t \right) = c{t^4}$, where $c$ is to be determined.
Since the path ${\bf{r}}\left( t \right) = \left( {t,g\left( t \right)} \right)$ passes through $\left( {1,2} \right)$ at $t=1$, we have
${\bf{r}}\left( 1 \right) = \left( {1,c} \right)) = \left( {1,2} \right)$
So, $c=2$. Thus, the path is ${\bf{r}}\left( t \right) = \left( {t,2{t^4}} \right)$.
