Question Number 188083 by Kalebwizeman last updated on 25/Feb/23
$${determine}\:{the}\:{value}\:{of}\:{b}\:{for}\:{which}\: \\ $$$$\:\:\:{y}=\frac{−{x}}{\mathrm{3}}\:+{b}\:\:{meets}\:{the}\:{graph}\:{of} \\ $$$$\:{y}^{\mathrm{2}} ={x}^{\mathrm{3}} \:\:{orthogonally} \\ $$
Commented by a.lgnaoui last updated on 27/Feb/23
 b−(x_0 /3) =(√x_0 ^3 ) (2)(intersect) (1)⇒((10)/3)x_0 =b+(c/3) x_0 =((10)/3)b+((10c)/9) b=((10)/3)x_0 −(c/3) ⇒ b−(x_0 /3)=3x_0 −(c/3)=f(x_0 ) ou 3x_0 +(b−((10x_0 )/3))=tangente 1)− b−(x_0 /3)+(c/3)=3x_0 2 )− 3x_0 −(c/3)=(√x_0 ^3 ) x^3 −9x^2 +2cx−((c/3))^2 =0 (a completer...) A suivre...................](https://www.tinkutara.com/question/Q188205.png)
$$\mathrm{tan}\:\theta=\frac{{y}}{{x}}=\frac{{dy}}{{dx}}\:\:\:{y}={x}\frac{{dy}}{{dx}} \\ $$$$\:\mathrm{3}=\frac{{b}−\frac{{x}}{\mathrm{3}}+\frac{{c}}{\mathrm{3}}}{{x}}\:\:\:\:\:\left[\left(\mathrm{1}\right)\right]\left({meme}\:{tengente}\right) \\ $$$$\:{b}−\frac{{x}_{\mathrm{0}} }{\mathrm{3}}\:\:=\sqrt{{x}_{\mathrm{0}} ^{\mathrm{3}} }\:\:\:\:\:\:\:\left(\mathrm{2}\right)\left({intersect}\right) \\ $$$$\left(\mathrm{1}\right)\Rightarrow\frac{\mathrm{10}}{\mathrm{3}}{x}_{\mathrm{0}} ={b}+\frac{{c}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:{x}_{\mathrm{0}} =\frac{\mathrm{10}}{\mathrm{3}}{b}+\frac{\mathrm{10}{c}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{b}=\frac{\mathrm{10}}{\mathrm{3}}{x}_{\mathrm{0}} −\frac{{c}}{\mathrm{3}}\:\:\:\:\:\Rightarrow \\ $$$$\:\:\:\:{b}−\frac{{x}_{\mathrm{0}} }{\mathrm{3}}=\mathrm{3}{x}_{\mathrm{0}} −\frac{{c}}{\mathrm{3}}={f}\left({x}_{\mathrm{0}} \right) \\ $$$${ou}\:\:\:\:\mathrm{3}{x}_{\mathrm{0}} +\left({b}−\frac{\mathrm{10}{x}_{\mathrm{0}} }{\mathrm{3}}\right)={tangente} \\ $$$$\left.\:\mathrm{1}\right)−\:\:{b}−\frac{{x}_{\mathrm{0}} }{\mathrm{3}}+\frac{{c}}{\mathrm{3}}=\mathrm{3}{x}_{\mathrm{0}} \:\:\: \\ $$$$\left.\mathrm{2}\:\right)−\:\:\:\mathrm{3}{x}_{\mathrm{0}} −\frac{{c}}{\mathrm{3}}=\sqrt{{x}_{\mathrm{0}} ^{\mathrm{3}} }\: \\ $$$${x}^{\mathrm{3}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{2}{cx}−\left(\frac{{c}}{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}\:{completer}…\right) \\ $$$${A}\:\:{suivre}………………. \\ $$
Commented by a.lgnaoui last updated on 26/Feb/23
Commented by Kalebwizeman last updated on 26/Feb/23
$${thank}\:{you} \\ $$$${what}\:{is}\:{now}\:{the}\:{value}\:{of}\:{b}? \\ $$
Answered by mr W last updated on 27/Feb/23
$${at}\:{intersection}\:{point}: \\ $$$$\mathrm{2}{y}\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3} \\ $$$$\Rightarrow\mathrm{2}{y}×\mathrm{3}=\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\Rightarrow{y}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}>\mathrm{0} \\ $$$${on}\:{the}\:{other}\:{side} \\ $$$${y}^{\mathrm{2}} ={x}^{\mathrm{3}} \\ $$$$\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} ={x}^{\mathrm{3}} \\ $$$$\Rightarrow{x}=\mathrm{4} \\ $$$$\Rightarrow{y}=\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{8} \\ $$$$\mathrm{8}=−\frac{\mathrm{4}}{\mathrm{3}}+{b} \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}}{\mathrm{3}}+\mathrm{8}=\frac{\mathrm{28}}{\mathrm{3}}\:\checkmark \\ $$
Commented by mr W last updated on 27/Feb/23
