Question

Find a unit vector that is parallel to the line tangent to the parabola y = x2 at the point (5, 25).

Answer 1

The line tangent to $y=x^2$ at the point (5, 25) as slope equal to $y'$ when $x=5$:

$y=x^2\implies y'=2x\implies\mathrm{slope}=10$

Then the tangent has equation

$y-25=10(x-5)\implies y=10x-25$

But only the slope is important here; any vector $t(1,10)$ is parallel to this line. Then a unit vector would be obtained by dividing this vector by its norm. Let $t=1$ for simplicity; then the unit vector is

$\dfrac{(1,10)}{\sqrt{1^2+10^2}}=\left(\dfrac1{\sqrt{101}},\dfrac{10}{\sqrt{101}}\right)$