Question Number 98925 by ajfour last updated on 17/Jun/20
$${If}\:{the}\:{curve}\:{shown}\:{below}\:{has}\:{the}\: \\ $$$${equation},\:\:{y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{bx}−{c}\right) \\ $$$${then}\:{find}\:\:{q}/{p}\:\:{in}\:{terms}\:{of}\:{b}\:{and}\:{c}. \\ $$
Commented by ajfour last updated on 17/Jun/20
Answered by mr W last updated on 17/Jun/20
$${eqn}.\:{of}\:{tangent}\:{line}: \\ $$$$\frac{{x}}{{p}}+\frac{{y}}{{q}}=\mathrm{1} \\ $$$$\Rightarrow{y}={q}\left(\mathrm{1}−\frac{{x}}{{p}}\right) \\ $$$${q}\left(\mathrm{1}−\frac{{x}}{{p}}\right)=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{bx}−{c}\right) \\ $$$$\Rightarrow{x}^{\mathrm{3}} −{bx}−{c}+\frac{{q}}{{p}}=\mathrm{0} \\ $$$${it}\:{should}\:{have}\:{one}\:{single}\:{and}\:{one} \\ $$$${double}\:{root}. \\ $$$${x}^{\mathrm{3}} −{bx}−{c}+\frac{{q}}{{p}}=\left({x}−{u}\right)\left({x}−{v}\right)^{\mathrm{2}} \\ $$$${u}+\mathrm{2}{v}=\mathrm{0} \\ $$$$\mathrm{2}{uv}+{v}^{\mathrm{2}} =−{b} \\ $$$${uv}^{\mathrm{2}} ={c}−\frac{{q}}{{p}} \\ $$$$\Rightarrow−\mathrm{4}{v}^{\mathrm{2}} +{v}^{\mathrm{2}} =−{b}\:\Rightarrow{v}^{\mathrm{2}} =\frac{{b}}{\mathrm{3}} \\ $$$$\Rightarrow−\mathrm{2}{v}^{\mathrm{3}} ={c}−\frac{{q}}{{p}}\:\Rightarrow{v}^{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{q}}{{p}}−{c}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{{q}}{{p}}−{c}\right)^{\mathrm{2}} =\left(\frac{{b}}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\frac{{q}}{{p}}={c}+\frac{\mathrm{2}{b}}{\mathrm{3}}\sqrt{\frac{{b}}{\mathrm{3}}} \\ $$
Commented by mr W last updated on 17/Jun/20
Commented by ajfour last updated on 17/Jun/20
$${Thanks}\:{Sir},\:{perfect}! \\ $$