Question Number 216139 by mr W last updated on 28/Jan/25
Commented by mr W last updated on 28/Jan/25
$${find}\:\left({AP}+{PQ}\right)_{{min}} =? \\ $$
Commented by Ghisom last updated on 28/Jan/25
$$\mathrm{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{6} \\ $$
Commented by mr W last updated on 28/Jan/25
$${i}\:{think}\:{too}. \\ $$
Answered by mahdipoor last updated on 28/Jan/25
$${if}\:{P}\:=\left({x},{ax}^{\mathrm{2}} \right)\:{is}\:{fixed}\:\Rightarrow \\ $$$${QP}\:{is}\:{min}\:{when}\:{P}\:,\:{Q}\:,\:{B}\:{in}\:{one}\:{line} \\ $$$$\Rightarrow\:{QP}={PB}−{QB}={PB}−{r} \\ $$$${s}={AP}+{PQ}= \\ $$$$\sqrt{{x}^{\mathrm{2}} +\left({ax}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({ax}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} }−\mathrm{2} \\ $$$$\frac{{ds}}{{dx}}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{3}\:\Rightarrow\:{s}_{{min}} =\mathrm{8}\: \\ $$
Answered by mr W last updated on 29/Jan/25
Commented by mr W last updated on 29/Jan/25
$${A}={focus}\:{of}\:{parabola} \\ $$$${BQ}={r}=\mathrm{2} \\ $$$$\left({AP}+{PQ}\underset{{min}} {\right)}\Leftrightarrow\left({AP}+{PQ}+{QB}\right)_{{min}} \\ $$$$\left({AP}+{PQ}+{QB}\right)_{{min}} =\left({AP}+{PB}\right)_{{min}} =\left({CP}+{PB}\right)_{{min}} =\mathrm{6}+\mathrm{2}=\mathrm{8} \\ $$$$\left({AP}+{PQ}\right)_{{min}} =\mathrm{8}−{r}=\mathrm{6}\:\checkmark \\ $$