Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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.

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

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.

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

Commented by mrW1 last updated on 23/Dec/17

Commented by ajfour last updated on 24/Dec/17

thanks Sir!

$${thanks}\:{Sir}! \\ $$

Commented by mrW1 last updated on 23/Dec/17

Terms of Service

Privacy Policy

Contact: info@tinkutara.com