Question Number 55613 by ajfour last updated on 27/Feb/19
Commented by ajfour last updated on 27/Feb/19
$${red}\:{curve}\:{is}\:\:\:{y}={x}^{\mathrm{3}} +{px}+{q} \\ $$$${blue}\:{line}\:{is}\:{its}\:{tangent}\:{at}\:{its} \\ $$$${largest}\:{root}. \\ $$$${After}\:{rotation}\:{of}\:{the}\:{curve} \\ $$$${about}\:{this}\:{root}\:{point}\:{P}\:\:{the}\:{same} \\ $$$${tangent}\:{if}\:{rotated}\:{together}\:{with} \\ $$$${the}\:{curve}\:{now}\:{coincides}\:{with}\:{the} \\ $$$${x}-{axis}.\:{Find}\:{the}\:{equation}\:{of}\:{the} \\ $$$${new}\:{curve},\:{hence}\:\alpha. \\ $$
Commented by mr W last updated on 28/Feb/19
Commented by mr W last updated on 28/Feb/19
![when the curve y=f(x) is rotated about point P(h,k) by an angle θ (+↷), then the eqn. of the new curve is y_1 =f(x_1 ) with y_1 =(x−h)sin θ+(y−k)cos θ+k x_1 =(x−h)cos θ−(y−k)sin θ+h or F(x,y)=(x−h)sin θ+(y−k)cos θ+k−f[(x−h)cos θ−(y−k)sin θ+h]=0](https://www.tinkutara.com/question/Q55617.png)
$${when}\:{the}\:{curve}\:{y}={f}\left({x}\right)\:{is}\:{rotated}\:{about} \\ $$$${point}\:{P}\left({h},{k}\right)\:{by}\:{an}\:{angle}\:\theta\:\left(+\curvearrowright\right), \\ $$$${then}\:{the}\:{eqn}.\:{of}\:{the}\:{new}\:{curve}\:{is} \\ $$$${y}_{\mathrm{1}} ={f}\left({x}_{\mathrm{1}} \right)\:{with} \\ $$$${y}_{\mathrm{1}} =\left({x}−{h}\right)\mathrm{sin}\:\theta+\left({y}−{k}\right)\mathrm{cos}\:\theta+{k} \\ $$$${x}_{\mathrm{1}} =\left({x}−{h}\right)\mathrm{cos}\:\theta−\left({y}−{k}\right)\mathrm{sin}\:\theta+{h} \\ $$$${or} \\ $$$${F}\left({x},{y}\right)=\left({x}−{h}\right)\mathrm{sin}\:\theta+\left({y}−{k}\right)\mathrm{cos}\:\theta+{k}−{f}\left[\left({x}−{h}\right)\mathrm{cos}\:\theta−\left({y}−{k}\right)\mathrm{sin}\:\theta+{h}\right]=\mathrm{0} \\ $$
Commented by mr W last updated on 28/Feb/19
Commented by ajfour last updated on 28/Feb/19
$${Yes}\:{Sir},\:{thank}\:{you}.\:{I}\:{was}\:{mistaken} \\ $$$${in}\:{thinking}\:{that}\:{this}\:{transformation} \\ $$$${might}\:{help}\:{determining}\:{the}\:{root}. \\ $$