I have to find the shortest distance between the line $x - y - 2= 0$ and the parabola $x^2 = y.$
This is my solution but I'm not sure if it is right. Point on $y=x^2$ is with coordinates $(t^2, t)$. Distance between the point $(t^2, t)$ from the line x - y - 2 = 0 is: $p = \left | \frac{t^2-t-2}{\sqrt{A^2+B^2}} \right |$ $$p=\left | \frac{t^2-t-2}{\sqrt{1^2+(-1)^2}} \right |$$ $$p=\left | \frac{t^2-t-2}{\sqrt{2}} \right |$$ p is the smallest when the first derivative is equal to 0. the first derivative of p is $$\sqrt{2}t-\frac{\sqrt{2}}{2}=0.$$ It means that t is equal to $\frac{1}{2}.$ So, the answer is $$\frac{1}{2}^{2}-\frac{1}{4}-2= \frac{9}{4\sqrt{2}}$$. Is this right?
HINT.-(1)Take a point $P_0=(x_0,x_o^2)$ in which the tangent to the parabola be parallel to the line $y=x-2)$. The point is $(0.5,0.25)$ (because in the tangent $y=2x_0x-x_0^2$ we need $2x_0=1$).
(2)Find the distance between $y=x-2$ and $y=x-\dfrac14$ (for example the perpendicular line from $(0.5,0.25)$ to $y=x-2$ is $y=-x+\dfrac14$, then find the distance from $P_0$ to the intersection of lines $y=-x+\dfrac14$ and $y=x-2$.