S6.12.3.35 Find the unit vectors

[karush]

$\tiny{s6.12.3.35}\\$
35. Find the unit vectors that are parallel to the tangent line to the
parabola $y = x^2$ at the point $(2,4)$.
\begin{align}
\displaystyle
y'&=2x
\end{align}
the book answer to this is
$\pm\left(i+4j)/\sqrt{17}\right)$
but don't see how they got this?

 

[Greg]

The equation of the tangent line there is y = 4x - 4. Slope of 4, so it runs for 1 and rises 4, giving us i + 4j. The magnitude of this vector is

$$\sqrt{1^2+4^2}=\sqrt{17}$$

Got it now?

 

[karush]

$\tiny{s6.12.3.35}\\$
35. Find the unit vectors that are parallel to the tangent line to the
parabola $y = x^2$ at the point $(2,4)$.
\begin{align}
\displaystyle
y'&=2x\\
\textsf{tangent line at (2,4) is}\\
y&=4(x-2)+4=4x-4\\
\sqrt{1^2+4^2}&=\sqrt{17}\\
\textsf{thus}\\
V_{35}&=\pm\left(i+4j)/\sqrt{17}\right)
\end{align}