Question Number 126605 by bramlexs22 last updated on 22/Dec/20
$$\:{Let}\:{P}\:{be}\:{point}\:{on}\:{the}\:{graph}\: \\ $$$${of}\:{a}\:{straight}\:{line}\:{y}=\mathrm{2}{x}−\mathrm{3}\:{and}\:{Q} \\ $$$${be}\:{a}\:{point}\:{on}\:{the}\:{graph}\:{of}\:{a}\:{parabola} \\ $$$${y}={x}^{\mathrm{2}} +{x}+\mathrm{1}\:.{Find}\:{the}\:{shortest}\: \\ $$$${distance}\:{between}\:{P}\:{and}\:{Q}\:. \\ $$
Answered by liberty last updated on 22/Dec/20
$${let}\:{y}=\mathrm{2}{x}+{p}\:{is}\:{tangent}\:{line}\:{at}\:{curve}\: \\ $$$${y}={x}^{\mathrm{2}} +{x}+\mathrm{1}\:\Rightarrow\:{we}\:{get}\:\mathrm{2}{x}+\mathrm{1}=\mathrm{2}\:;\:{x}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${the}\:{point}\:{Q}\:\left(\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{7}}{\mathrm{4}}\right)\:{and}\:{p}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${thus}\:\mid{PQ}\mid_{{min}} \:=\:\mid\:\frac{\frac{\mathrm{3}}{\mathrm{4}}−\left(−\mathrm{3}\right)}{\:\sqrt{\mathrm{5}}}\mid\:=\:\frac{\mathrm{15}}{\mathrm{4}\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{3}}{\mathrm{4}}\sqrt{\mathrm{5}} \\ $$