Question Number 31113 by NECx last updated on 02/Mar/18
$${Two}\:{lines}\:{through}\:{the}\:{point}\:\left(\mathrm{1},−\mathrm{3}\right) \\ $$$${are}\:{tamgent}\:{to}\:{the}\:{curve}\:{y}={x}^{\mathrm{2}} . \\ $$$${Find}\:{the}\:{equation}\:{of}\:{these}\:{two} \\ $$$${lines}\:{and}\:{make}\:{a}\:{sketch}\:{to}\:{verify} \\ $$$${your}\:{results}. \\ $$
Answered by Tinkutara last updated on 02/Mar/18
Commented by Tinkutara last updated on 02/Mar/18
$${Let}\:{equation}\:{of}\:{tangent}\:{is}\:{y}={mx}+{c} \\ $$$${It}\:{touches}\:{y}={x}^{\mathrm{2}} \:{so}\:{D}=\mathrm{0} \\ $$$${mx}+{c}={x}^{\mathrm{2}} \\ $$$${D}=\mathrm{0}\:{gives}\:{c}=\frac{−{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${y}={mx}−\frac{{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${It}\:{passes}\:{through}\:\left(\mathrm{1},−\mathrm{3}\right) \\ $$$$−\mathrm{3}={m}−\frac{{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${m}=−\mathrm{2},\mathrm{6} \\ $$$${Equations}\:{of}\:{line}\:{are} \\ $$$${y}=−\mathrm{2}{x}−\mathrm{1} \\ $$$${y}=\mathrm{6}{x}−\mathrm{9} \\ $$
Commented by NECx last updated on 03/Mar/18
$${I}'{m}\:{most}\:{grateful}\:{sir}.{Thanks}. \\ $$
Answered by mrW2 last updated on 02/Mar/18
$${eqn}.\:{of}\:{line}\:{through}\:{point}\:\left(\mathrm{1},−\mathrm{3}\right): \\ $$$${y}+\mathrm{3}={m}\left({x}−\mathrm{1}\right) \\ $$$${or}\:{y}={m}\left({x}−\mathrm{1}\right)−\mathrm{3} \\ $$$$ \\ $$$${intersection}: \\ $$$${y}={m}\left({x}−\mathrm{1}\right)−\mathrm{3}={x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{mx}+\left({m}+\mathrm{3}\right)=\mathrm{0} \\ $$$${D}={m}^{\mathrm{2}} −\mathrm{4}\left({m}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({m}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow{m}−\mathrm{2}=\pm\mathrm{4} \\ $$$$\Rightarrow{m}=\mathrm{6}\:{or}\:−\mathrm{2} \\ $$$$ \\ $$$${Eqn}.\:{of}\:{lines}: \\ $$$${y}=\mathrm{6}\left({x}−\mathrm{1}\right)−\mathrm{3}\Rightarrow{y}=\mathrm{6}{x}−\mathrm{9} \\ $$$${y}=−\mathrm{2}\left({x}−\mathrm{1}\right)−\mathrm{3}\Rightarrow{y}=−\mathrm{2}{x}−\mathrm{1} \\ $$
Commented by NECx last updated on 03/Mar/18
$${Thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$
Commented by NECx last updated on 02/Mar/18
$${and}\:{what}\:{does}\:{D}\:{represents}\:\:? \\ $$
Commented by mrW2 last updated on 02/Mar/18
$${If}\:{the}\:{eqn}.\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{should} \\ $$$${have}\:{only}\:{one}\:{solution},\:{then} \\ $$$${D}={b}^{\mathrm{2}} −\mathrm{4}{ac}\:{must}\:{equal}\:{to}\:{zero}. \\ $$
Commented by NECx last updated on 02/Mar/18
$${why}\:{is}\:{D}=\mathrm{0}\:? \\ $$
Commented by mrW2 last updated on 02/Mar/18
