Question Number 42207 by Rio Michael last updated on 20/Aug/18
$${Find}\:{the}\:{equation}\:{of}\:{tangent}\:{and}\:{normal}\:{to}\:\:{the}\:{curve}\:{y} \\ $$$${given}\:{by}\:\:\:{y}\:=\:{x}^{\mathrm{3}} \:+\:\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{7}\:. \\ $$
Answered by MJS last updated on 20/Aug/18
$${f}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} +\mathrm{7} \\ $$$$\mathrm{tangent}\:\mathrm{in}\:\begin{pmatrix}{{p}}\\{{f}\left({p}\right)}\end{pmatrix} \\ $$$${y}={kx}+{d}\:\Rightarrow\:{d}={y}−{kx} \\ $$$${k}={f}'\left({p}\right)=\mathrm{3}{p}^{\mathrm{2}} +\mathrm{6}{p} \\ $$$${d}={f}\left({p}\right)−{f}'\left({p}\right){p}=−\mathrm{2}{p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7} \\ $$$$ \\ $$$${t}:\:{y}=\left(\mathrm{3}{p}^{\mathrm{2}} +\mathrm{6}{p}\right){x}−\mathrm{2}{p}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7} \\ $$$$ \\ $$$$\mathrm{normal}\:\mathrm{in}\:\begin{pmatrix}{{p}}\\{{f}\left({p}\right)}\end{pmatrix} \\ $$$${y}=−\frac{\mathrm{1}}{{k}}{x}+{d}\:\Rightarrow\:{d}={y}+\frac{\mathrm{1}}{{k}}{x} \\ $$$${k}=\mathrm{same}\:\mathrm{as}\:\mathrm{above}={f}'\left({p}\right) \\ $$$${d}={f}\left({p}\right)+\frac{\mathrm{1}}{{f}'\left({p}\right)}{p}={p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7}+\frac{{p}}{\mathrm{3}{p}^{\mathrm{2}} +\mathrm{6}{p}}= \\ $$$$={p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7}+\frac{\mathrm{1}}{\mathrm{3}{p}+\mathrm{6}}=\frac{\left({p}^{\mathrm{3}} +\mathrm{3}{p}^{\mathrm{2}} +\mathrm{7}\right)\left(\mathrm{3}{p}+\mathrm{6}\right)+\mathrm{1}}{\mathrm{3}{p}+\mathrm{6}}= \\ $$$$=\frac{\mathrm{3}{p}^{\mathrm{4}} +\mathrm{15}{p}^{\mathrm{3}} +\mathrm{18}{p}^{\mathrm{2}} +\mathrm{21}{p}+\mathrm{43}}{\mathrm{3}{p}+\mathrm{6}} \\ $$$$ \\ $$$${n}:\:{y}=\frac{\mathrm{1}}{\mathrm{3}\left({p}+\mathrm{2}\right)}\left(−{x}+\mathrm{3}{p}^{\mathrm{4}} +\mathrm{15}{p}^{\mathrm{3}} +\mathrm{18}{p}^{\mathrm{2}} +\mathrm{21}{p}+\mathrm{43}\right) \\ $$