Question Number 7900 by tawakalitu last updated on 23/Sep/16
$${Find}\:{an}\:{equation}\:{of}\:{the}\:{tangent}\:{line}\:{to}\:{the}\:{curve}\: \\ $$$${y}\:=\:{tan}^{\mathrm{2}} {x}\:\:\:{at}\:\:{the}\:{point}\:\:\left(\frac{\pi}{\mathrm{3}},\:\mathrm{3}\right) \\ $$
Commented by sou1618 last updated on 24/Sep/16
$${l}:{tangent}\:{line} \\ $$$$\mathrm{3}={tan}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{3}}\right) \\ $$$${so}\:{point}\left(\pi/\mathrm{3},\mathrm{3}\right)\:{is}\:{on}\:{l} \\ $$$$ \\ $$$${y}\:'=\mathrm{2}{tan}\left({x}\right)\centerdot\left({tan}\:{x}\right)' \\ $$$${y}\:'=\mathrm{2}{tan}\left({x}\right)\centerdot\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}} \\ $$$${l}:{y}−\mathrm{3}=\mathrm{2}{tan}\left(\pi/\mathrm{3}\right)\centerdot\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\pi/\mathrm{3}\right)}\left({x}−\pi/\mathrm{3}\right) \\ $$$${l}:{y}=\mathrm{2}\sqrt{\mathrm{3}}\centerdot\mathrm{4}\left({x}−\pi/\mathrm{3}\right)+\mathrm{3} \\ $$$${l}:{y}=\mathrm{8}\sqrt{\mathrm{3}}{x}+\mathrm{3}−\frac{\mathrm{8}\pi\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${l}:{y}=\mathrm{8}\sqrt{\mathrm{3}}{x}+\mathrm{3}−\frac{\mathrm{8}\pi}{\sqrt{\mathrm{3}}}\:. \\ $$$$ \\ $$
Commented by tawakalitu last updated on 24/Sep/16
$${Thank}\:{you}\:{so}\:{much}.\: \\ $$
Answered by prakash jain last updated on 02/Oct/16
$$\mathrm{see}\:\mathrm{comments} \\ $$