Question-26256

Question Number 26256 by ajfour last updated on 23/Dec/17

Commented by mrW1 last updated on 23/Dec/17

$${I}\:{have}\:{posted}\:{an}\:{additional}\:{image}. \\ $$$${I}\:{tried}\:{to}\:{move}\:{and}\:{rotate}\:{the}\:{coordinate} \\ $$$${system},\:{not}\:{the}\:{curve}.\:{The}\:{result} \\ $$$${is}\:{the}\:{same}. \\ $$

Answered by mrW1 last updated on 24/Dec/17

M(x_1 ,y_1 ) and M_2 (x_2 ,y_2 ) y_1 =Ax_1 ^2 tan 60°=(y_1 /(−x_1 ))=−Ax_1 ⇒Ax_1 =−(√3) x_1 ^2 +y_1 ^2 =a^2 x_1 ^2 +A^2 x_1 ^4 =a^2 x_1 ^2 +3x_1 ^2 =a^2 ⇒x_1 =−(a/2) ⇒A=((2(√3))/a) ⇒y_1 =(((√3)a)/2) Eqn. of M_1 M_2 : y−y_1 =tan 30° (x−x_1 ) y_2 −y_1 =(1/( (√3)))(x_2 −x_1 ) ((2(√3))/a) x_2 ^2 −(a/2)(√3)=(1/( (√3)))(x_2 +(a/2)) 6x_2 ^2 −ax_2 −2a^2 =0 (2x_2 +a)(3x_2 −2a)=0 x_2 =(2/3)a y_2 =((2(√3))/a)×((4a^2 )/9)=((8(√3))/9)a D=∣M_1 M_2 ∣ D^2 =(x_1 −x_2 )^2 +(y_1 −y_2 )^2 =(((2a)/3)+(a/2))^2 +(((8(√3)a)/9)−(((√3)a)/2))^2 =(((7a)/6))^2 +3(((7a)/(18)))^2 =((49a^2 )/(27)) ⇒D=((7a)/(3(√3))) Point Q(0,y_Q ) y_Q =D=((7a)/(3(√3)))

$${M}\left({x}_{\mathrm{1}} ,{y}_{\mathrm{1}} \right)\:{and}\:{M}_{\mathrm{2}} \left({x}_{\mathrm{2}} ,{y}_{\mathrm{2}} \right) \\ $$$${y}_{\mathrm{1}} ={Ax}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\mathrm{60}°=\frac{{y}_{\mathrm{1}} }{−{x}_{\mathrm{1}} }=−{Ax}_{\mathrm{1}} \\ $$$$\Rightarrow{Ax}_{\mathrm{1}} =−\sqrt{\mathrm{3}} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +{y}_{\mathrm{1}} ^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +{A}^{\mathrm{2}} {x}_{\mathrm{1}} ^{\mathrm{4}} ={a}^{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{3}{x}_{\mathrm{1}} ^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =−\frac{{a}}{\mathrm{2}} \\ $$$$\Rightarrow{A}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{a}} \\ $$$$\Rightarrow{y}_{\mathrm{1}} =\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}} \\ $$$$ \\ $$$${Eqn}.\:{of}\:{M}_{\mathrm{1}} {M}_{\mathrm{2}} : \\ $$$${y}−{y}_{\mathrm{1}} =\mathrm{tan}\:\mathrm{30}°\:\left({x}−{x}_{\mathrm{1}} \right) \\ $$$${y}_{\mathrm{2}} −{y}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right) \\ $$$$ \\ $$$$\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{a}}\:{x}_{\mathrm{2}} ^{\mathrm{2}} −\frac{{a}}{\mathrm{2}}\sqrt{\mathrm{3}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({x}_{\mathrm{2}} +\frac{{a}}{\mathrm{2}}\right) \\ $$$$\mathrm{6}{x}_{\mathrm{2}} ^{\mathrm{2}} −{ax}_{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{2}{x}_{\mathrm{2}} +{a}\right)\left(\mathrm{3}{x}_{\mathrm{2}} −\mathrm{2}{a}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{2}}{\mathrm{3}}{a} \\ $$$${y}_{\mathrm{2}} =\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{a}}×\frac{\mathrm{4}{a}^{\mathrm{2}} }{\mathrm{9}}=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{9}}{a} \\ $$$${D}=\mid{M}_{\mathrm{1}} {M}_{\mathrm{2}} \mid \\ $$$${D}^{\mathrm{2}} =\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{1}} −{y}_{\mathrm{2}} \right)^{\mathrm{2}} =\left(\frac{\mathrm{2}{a}}{\mathrm{3}}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{8}\sqrt{\mathrm{3}}{a}}{\mathrm{9}}−\frac{\sqrt{\mathrm{3}}{a}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{7}{a}}{\mathrm{6}}\right)^{\mathrm{2}} +\mathrm{3}\left(\frac{\mathrm{7}{a}}{\mathrm{18}}\right)^{\mathrm{2}} =\frac{\mathrm{49}{a}^{\mathrm{2}} }{\mathrm{27}} \\ $$$$\Rightarrow{D}=\frac{\mathrm{7}{a}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${Point}\:{Q}\left(\mathrm{0},{y}_{{Q}} \right) \\ $$$${y}_{{Q}} ={D}=\frac{\mathrm{7}{a}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

Commented by mrW1 last updated on 24/Dec/17

$${you}\:{are}\:{right}.\:{I}\:{have}\:{corrected}. \\ $$$${with}\:{my}\:{method}\:{I}\:{intented}\:{only}\:{to} \\ $$$${find}\:{the}\:{position}\:{of}\:{point}\:{Q}. \\ $$

Commented by mrW1 last updated on 23/Dec/17