Question Number 124133 by bramlexs22 last updated on 01/Dec/20
$${Given}\:{equation}\:{of}\:{tangent}\:{line} \\ $$$${of}\:{the}\:{curve}\:{y}\:=\:\frac{{b}}{{x}^{\mathrm{2}} }\:{at}\:{point}\:\left({x},{y}\right) \\ $$$${is}\:{bx}−\mathrm{4}{y}=−\mathrm{21}.\:{The}\:{value}\:{of}\:{b}\:=? \\ $$
Answered by mr W last updated on 01/Dec/20
$$\frac{{dy}}{{dx}}=−\frac{\mathrm{2}{b}}{{x}^{\mathrm{3}} } \\ $$$${tangent}\:{at}\:{point}\:\left({p},\frac{{b}}{{p}^{\mathrm{2}} }\right) \\ $$$${y}=−\frac{\mathrm{2}{b}}{{p}^{\mathrm{3}} }\left({x}−{p}\right)+\frac{{b}}{{p}^{\mathrm{2}} } \\ $$$$\Rightarrow{bx}+\frac{{p}^{\mathrm{3}} }{\mathrm{2}}{y}=\frac{\mathrm{3}}{\mathrm{2}}{pb} \\ $$$$\frac{{p}^{\mathrm{3}} }{\mathrm{2}}=−\mathrm{4}\:\Rightarrow{p}=−\mathrm{2} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}×\left(−\mathrm{2}\right){b}=−\mathrm{21}\:\Rightarrow{b}=\mathrm{7} \\ $$
Commented by mr W last updated on 01/Dec/20
Commented by I want to learn more last updated on 01/Dec/20
$$\mathrm{Sir}\:\mathrm{mrW},\:\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{with}\:\mathrm{permutation}\:\mathrm{on}\:\mathrm{Q124149} \\ $$
Answered by liberty last updated on 01/Dec/20
$${gradient}\:{of}\:{tangent}\:{line}\:\Rightarrow\:\frac{{b}}{\mathrm{4}}\:=\:−\frac{\mathrm{2}{b}}{{x}^{\mathrm{3}} }\: \\ $$$$\Rightarrow\begin{cases}{{x}^{\mathrm{3}} =−\mathrm{8}\:;\:{x}=−\mathrm{2}\:\wedge{y}\:=\:\frac{{b}}{\mathrm{4}}}\\{\Rightarrow−\mathrm{2}{b}−\mathrm{4}\left(\frac{{b}}{\mathrm{4}}\right)=−\mathrm{21}\:;\:−\mathrm{3}{b}=−\mathrm{21}\:;\:{b}=\mathrm{7}}\end{cases} \\ $$