min distance from the parabola to the line asked by Kl liu in the Department of Mathematics Credit to teacher Imad Zak

Exercise:

What is the min distance from the line $\left( d \right)\,\,\,y=2x-5$ and the parabola $\left( c \right)\,\,\,y={{x}^{2}}+x+1$

Solution: Let $\alpha \in {{\mathbb{R}}^{+}}$ and $M\in \left( c \right)$ then $M\left( \alpha ,{{\alpha }^{2}}+\alpha +1 \right)$  and we know that the min distance from line to

 a point is the perpendicular distance so observe that $\min \left\{ d\left( \left( d \right),\left( c \right) \right) \right\}=0$ occurs only when

 $\left( d \right)\cap \left( c \right)=\left\{ pt \right\}$ which is not the case

So $d\left( M,\left( d \right) \right)=\frac{\left| {{y}_{M}}-2{{x}_{M}}+5 \right|}{\sqrt{1+4}}=\frac{\left| {{\alpha }^{2}}+\alpha +1-2\alpha +5 \right|}{\sqrt{5}}=\frac{\left| {{\alpha }^{2}}-\alpha +6 \right|}{\sqrt{5}}=\frac{\left| f\left( \alpha  \right) \right|}{\sqrt{5}}$

hence $f'\left( \alpha  \right)=2\alpha -1$ so $f'\left( \alpha  \right)=0\Leftrightarrow \alpha =\frac{1}{2}$

so $f'>0$ when $\alpha >\frac{1}{2}$ and $f'<0$ when $\alpha <\frac{1}{2}$

thus $f$ has min at $\alpha =\frac{1}{2}$

hence $d\left( M,\left( d \right) \right)=\frac{\left| f\left( 1/2 \right) \right|}{\sqrt{5}}=\frac{23/4}{\sqrt{5}}=\frac{23}{4\sqrt{5}}$