Question Number 116301 by harckinwunmy last updated on 02/Oct/20 | ||
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{for}\:\mathrm{a} \\ $$$$\mathrm{given}\:\mathrm{line}\:\mathrm{to}\:\: \\ $$$$\left.\mathrm{1}\right)\:\mathrm{intersect}\:\mathrm{a}\:\mathrm{curve} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{be}\:\mathrm{a}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{a}\:\mathrm{curve} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{not}\:\mathrm{to}\:\mathrm{intersect}\:\mathrm{a}\:\mathrm{curve}\: \\ $$ | ||
Answered by Rio Michael last updated on 02/Oct/20 | ||
$$\left(\mathrm{1}\right)\:{b}^{\mathrm{2}} −\mathrm{4}{ac}\:>\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:{b}^{\mathrm{2}} −\mathrm{4}{ac}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:{b}^{\mathrm{2}} −\mathrm{4}{ac}\:<\:\mathrm{0} \\ $$ | ||
Answered by mr W last updated on 02/Oct/20 | ||
$${say}\:{the}\:{line}\:{is}\:{y}={ax}+{b} \\ $$$${the}\:{curve}\:{is}\:{y}={f}\left({x}\right) \\ $$$$\left(\mathrm{1}\right) \\ $$$${if}\:{f}\left({x}\right)−{ax}−{b}=\mathrm{0}\:{has}\:{at}\:{least}\:{one} \\ $$$${real}\:{root},\:{say}\:{the}\:{roots}\:{are}\:{r}_{{i}} ,\:{and}\: \\ $$$${f}\:'\left({r}_{{i}} \right)\neq{a} \\ $$$$ \\ $$$$\left(\mathrm{2}\right) \\ $$$${as}\:\left(\mathrm{1}\right)\:{but}\:{f}\:'\left({r}_{{i}} \right)={a} \\ $$$$ \\ $$$$\left(\mathrm{3}\right) \\ $$$${if}\:{g}\left({x}\right)−{ax}−{b}=\mathrm{0}\:{has}\:{no}\:{real}\:{root}. \\ $$ | ||