Question Number 83495 by john santu last updated on 03/Mar/20
$$\mathrm{closest}\:\mathrm{distance}\:\mathrm{point}\:\left(\mathrm{3},\mathrm{0}\right)\:\mathrm{to}\:\mathrm{curve}\: \\ $$$$\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}+\mathrm{4}\:? \\ $$
Commented by john santu last updated on 03/Mar/20
$$\mathrm{d}\:=\:\sqrt{\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{y}\right)^{\mathrm{2}} } \\ $$$$\mathrm{d}^{\mathrm{2}} \:=\:\left(\mathrm{x}−\mathrm{3}\right)^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{x}+\mathrm{4}+\mathrm{x}^{\mathrm{2}} −\mathrm{6x}+\mathrm{9}−\mathrm{d}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{13}−\mathrm{d}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{tangency}\:\bigtriangleup\:=\:\mathrm{25}−\mathrm{4}\left(\mathrm{13}−\mathrm{d}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\frac{\mathrm{25}}{\mathrm{4}}\:=\:\mathrm{13}−\mathrm{d}^{\mathrm{2}} \:\Rightarrow\:\mathrm{d}=\sqrt{\frac{\mathrm{52}−\mathrm{27}}{\mathrm{4}}\:}\:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by john santu last updated on 03/Mar/20
$$\mathrm{dear}\:\mathrm{Mr}\:\mathrm{W}.\:\mathrm{i}\:\mathrm{try}\:\mathrm{your}\:\mathrm{method}. \\ $$$$\mathrm{this}\:\mathrm{correct}? \\ $$
Commented by jagoll last updated on 03/Mar/20
$$\mathrm{good}\:\mathrm{sir} \\ $$
Commented by jagoll last updated on 03/Mar/20
$$\mathrm{x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{13}−\mathrm{d}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{tangency}\:\bigtriangleup\:=\:\mathrm{25}−\mathrm{4}\left(\mathrm{13}−\mathrm{d}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{13}−\mathrm{d}^{\mathrm{2}} \:=\:\frac{\mathrm{25}}{\mathrm{4}}\:\Leftrightarrow\:\mathrm{d}^{\mathrm{2}} \:=\:\frac{\mathrm{52}−\mathrm{25}}{\mathrm{4}} \\ $$$$\mathrm{d}\:=\:\sqrt{\frac{\mathrm{27}}{\mathrm{4}}}\:=\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$
Commented by john santu last updated on 03/Mar/20
$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$