Question Number 94773 by Hamida last updated on 20/May/20
Answered by prakash jain last updated on 20/May/20
$${slope}=\frac{{dy}}{{dx}}=\frac{{dy}/{dt}}{{dx}/{dt}}=\frac{\mathrm{8}{t}+\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{9}} \\ $$$${at}\:{t}=\mathrm{1}\:\mathrm{slopd}=\frac{\mathrm{9}}{\mathrm{12}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{point}\:\mathrm{on}\:\mathrm{curve}\:\mathrm{at}\:{t}=\mathrm{1} \\ $$$${y}=\mathrm{5},\:{x}=\mathrm{10} \\ $$$${tangent} \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{4}}{x}+{c}\:\:\left({i}\right) \\ $$$$\mathrm{it}\:\mathrm{pased}\:\mathrm{thru}\:\mathrm{point}\:\left(\mathrm{10},\mathrm{5}\right) \\ $$$${substitute}\:{in}\:\left({i}\right)\:{to}\:{vet}\:{c} \\ $$