Question Number 128521 by bramlexs22 last updated on 08/Jan/21
$$\mathrm{A}\:\mathrm{curve}\:\mathrm{has}\:\mathrm{equation}\:\mathrm{y}\:=\:\left(\mathrm{2}−\sqrt{\mathrm{x}}\:\right)^{\mathrm{4}} \\ $$$$\mathrm{The}\:\mathrm{normal}\:\mathrm{at}\:\mathrm{point}\:\mathrm{P}\left(\mathrm{1},\mathrm{1}\right)\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{normal}\:\mathrm{at}\:\mathrm{point}\:\mathrm{Q}\left(\mathrm{9},\mathrm{1}\right)\:\mathrm{intersect} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{point}\:\mathrm{R}.\:\mathrm{What}\:\mathrm{are}\:\mathrm{coordinates} \\ $$$$\mathrm{at}\:\mathrm{point}\:\mathrm{R}? \\ $$
Answered by mr W last updated on 08/Jan/21
$$\frac{{dy}}{{dx}}=\mathrm{4}\left(\mathrm{2}−\sqrt{{x}}\right)^{\mathrm{3}} \left(−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}}\right)=−\frac{\mathrm{2}\left(\mathrm{2}−\sqrt{{x}}\right)^{\mathrm{3}} }{\:\sqrt{{x}}} \\ $$$${at}\:{P}\left(\mathrm{1},\mathrm{1}\right): \\ $$$$\frac{{dy}}{{dx}}=−\mathrm{2} \\ $$$${eqn}.\:{of}\:{normal}: \\ $$$${y}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\mathrm{1}\right)=\frac{{x}+\mathrm{1}}{\mathrm{2}} \\ $$$${at}\:{Q}\left(\mathrm{9},\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${eqn}.\:{of}\:{normal}: \\ $$$${y}=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}\left({x}−\mathrm{9}\right)=\frac{−\mathrm{3}{x}+\mathrm{29}}{\mathrm{2}} \\ $$$${intersection}: \\ $$$${y}=\frac{{x}+\mathrm{1}}{\mathrm{2}}=\frac{−\mathrm{3}{x}+\mathrm{29}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{7} \\ $$$$\Rightarrow{y}=\mathrm{4} \\ $$$$\Rightarrow{R}\left(\mathrm{7},\mathrm{4}\right) \\ $$
Commented by mr W last updated on 08/Jan/21
Commented by bramlexs22 last updated on 08/Jan/21
$$\mathrm{great}\:\mathrm{sir}.\:\mathrm{thank}\:\mathrm{you} \\ $$