Answer
The tangent vector required is at $t = \frac{1}{\alpha } = \frac{1}{{100\sqrt 5 }}$, which is
${\bf{r}}'\left( {\frac{1}{{100\sqrt 5 }}} \right) = \left( {100\sqrt 5 ,200\sqrt 5 } \right)$
Work Step by Step
We have $y = {x^2}$. Since parametrization is not unique, let $x = \alpha t$, where $\alpha$ is a scalar. So, we may parametrize the curve by ${\bf{r}}\left( t \right) = \left( {\alpha t,{\alpha ^2}{t^2}} \right)$. The derivative is ${\bf{r}}'\left( t \right) = \left( {\alpha ,2{\alpha ^2}t} \right)$.
When the jet is at the point $\left( {1,1} \right)$, it has a speed of $500$ km/h. In our parametrization, the point $\left( {1,1} \right)$ corresponds to $t = \frac{1}{\alpha }$. So, at $t = \frac{1}{\alpha }$, $||{\bf{r}}'\left( {\frac{1}{\alpha }} \right)|| = 500$ km/h.
Since ${\bf{r}}'\left( t \right) = \left( {\alpha ,2{\alpha ^2}t} \right)$, so
$||{\bf{r}}'\left( t \right)|{|^2} = \left( {\alpha ,2{\alpha ^2}t} \right)\cdot\left( {\alpha ,2{\alpha ^2}t} \right)$
$||{\bf{r}}'\left( t \right)|{|^2} = {\alpha ^2} + 4{\alpha ^4}{t^2}$
$||{\bf{r}}'\left( t \right)|| = \sqrt {{\alpha ^2} + 4{\alpha ^4}{t^2}} $
Since $||{\bf{r}}'\left( {\frac{1}{\alpha }} \right)|| = 500$, so
$||{\bf{r}}'\left( {\frac{1}{\alpha }} \right)|| = \sqrt {{\alpha ^2} + 4{\alpha ^2}} = 500$
$5{\alpha ^2} = 250000$
$\alpha = \sqrt {50000} = 100\sqrt 5 $
Thus, the parametrization is
${\bf{r}}\left( t \right) = \left( {\alpha t,{\alpha ^2}{t^2}} \right) = \left( {100\sqrt 5 t,50000{t^2}} \right)$
The tangent vector is ${\bf{r}}'\left( t \right) = \left( {100\sqrt 5 ,100000t} \right)$. The tangent vector required is at $t = \frac{1}{\alpha } = \frac{1}{{100\sqrt 5 }}$, which is
${\bf{r}}'\left( {\frac{1}{{100\sqrt 5 }}} \right) = \left( {100\sqrt 5 ,\frac{{1000}}{{\sqrt 5 }}} \right)$
${\bf{r}}'\left( {\frac{1}{{100\sqrt 5 }}} \right) = \left( {100\sqrt 5 ,200\sqrt 5 } \right)$
