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Question-162309




Question Number 162309 by Mathematification last updated on 28/Dec/21
Commented by mr W last updated on 28/Dec/21
it seems that you and ms. tawa are  visiting the same college, since your  questions seem to be from the same  source (or teacher).
$${it}\:{seems}\:{that}\:{you}\:{and}\:{ms}.\:{tawa}\:{are} \\ $$$${visiting}\:{the}\:{same}\:{college},\:{since}\:{your} \\ $$$${questions}\:{seem}\:{to}\:{be}\:{from}\:{the}\:{same} \\ $$$${source}\:\left({or}\:{teacher}\right). \\ $$
Commented by Mathematification last updated on 28/Dec/21
The question is really from my friend, I shared   it in a whatsapp group, that was when m.s tawa  (whom i dont know) found the question.   I also saw the previous questions she posted here.
$${The}\:{question}\:{is}\:{really}\:{from}\:{my}\:{friend},\:{I}\:{shared}\: \\ $$$${it}\:{in}\:{a}\:{whatsapp}\:{group},\:{that}\:{was}\:{when}\:{m}.{s}\:{tawa} \\ $$$$\left({whom}\:{i}\:{dont}\:{know}\right)\:{found}\:{the}\:{question}.\: \\ $$$${I}\:{also}\:{saw}\:{the}\:{previous}\:{questions}\:{she}\:{posted}\:{here}. \\ $$
Commented by mr W last updated on 28/Dec/21
i see. thanks for clarifing!
$${i}\:{see}.\:{thanks}\:{for}\:{clarifing}! \\ $$
Answered by mr W last updated on 28/Dec/21
Commented by Mathematification last updated on 29/Dec/21
Please how did you know that the coordinate  connecting the parabola and the circumference   is (p, p^2 )      ??
$${Please}\:{how}\:{did}\:{you}\:{know}\:{that}\:{the}\:{coordinate} \\ $$$${connecting}\:{the}\:{parabola}\:{and}\:{the}\:{circumference}\: \\ $$$${is}\:\left({p},\:{p}^{\mathrm{2}} \right)\:\:\:\:\:\:?? \\ $$
Commented by mr W last updated on 28/Dec/21
P(p,p^2 )  tan θ=(dy/dx)=2x=2p  x_A =p+R sin θ  y_A =p^2 −R cos θ  (x_A −7)^2 +(y_A −3)^2 =(3+R)^2    ...(i)  (x_A −11)^2 +(y_A −3)^2 =(1+R)^2    ...(ii)  (i)−(ii):  4(2x_A −18)=2(2R+4)  ⇒x_A =(R/2)+10  (R/2)+10=p+R sin θ=p+((2pR)/( (√(1+4p^2 ))))  ⇒R=((2(10−p))/(((4p)/( (√(1+4p^2 ))))−1))  ⇒((R/2)+3)^2 +(p^2 −(R/( (√(1+4p^2 ))))−3)^2 =(3+R)^2   ⇒p≈4.0641  ⇒R≈12.0522
$${P}\left({p},{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\theta=\frac{{dy}}{{dx}}=\mathrm{2}{x}=\mathrm{2}{p} \\ $$$${x}_{{A}} ={p}+{R}\:\mathrm{sin}\:\theta \\ $$$${y}_{{A}} ={p}^{\mathrm{2}} −{R}\:\mathrm{cos}\:\theta \\ $$$$\left({x}_{{A}} −\mathrm{7}\right)^{\mathrm{2}} +\left({y}_{{A}} −\mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{3}+{R}\right)^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$$\left({x}_{{A}} −\mathrm{11}\right)^{\mathrm{2}} +\left({y}_{{A}} −\mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{1}+{R}\right)^{\mathrm{2}} \:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\mathrm{4}\left(\mathrm{2}{x}_{{A}} −\mathrm{18}\right)=\mathrm{2}\left(\mathrm{2}{R}+\mathrm{4}\right) \\ $$$$\Rightarrow{x}_{{A}} =\frac{{R}}{\mathrm{2}}+\mathrm{10} \\ $$$$\frac{{R}}{\mathrm{2}}+\mathrm{10}={p}+{R}\:\mathrm{sin}\:\theta={p}+\frac{\mathrm{2}{pR}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }} \\ $$$$\Rightarrow{R}=\frac{\mathrm{2}\left(\mathrm{10}−{p}\right)}{\frac{\mathrm{4}{p}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}−\mathrm{1}} \\ $$$$\Rightarrow\left(\frac{{R}}{\mathrm{2}}+\mathrm{3}\right)^{\mathrm{2}} +\left({p}^{\mathrm{2}} −\frac{{R}}{\:\sqrt{\mathrm{1}+\mathrm{4}{p}^{\mathrm{2}} }}−\mathrm{3}\right)^{\mathrm{2}} =\left(\mathrm{3}+{R}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{p}\approx\mathrm{4}.\mathrm{0641} \\ $$$$\Rightarrow{R}\approx\mathrm{12}.\mathrm{0522} \\ $$
Commented by mr W last updated on 29/Dec/21
Commented by Mathematification last updated on 29/Dec/21
Please sir, which textbook can overshadow this ??
$${Please}\:{sir},\:{which}\:{textbook}\:{can}\:{overshadow}\:{this}\:?? \\ $$
Commented by peter frank last updated on 29/Dec/21
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by mr W last updated on 29/Dec/21
this is about the application of   analytic geometry. i have no   special textbook to recommend.
$${this}\:{is}\:{about}\:{the}\:{application}\:{of}\: \\ $$$${analytic}\:{geometry}.\:{i}\:{have}\:{no}\: \\ $$$${special}\:{textbook}\:{to}\:{recommend}. \\ $$
Commented by Mathematification last updated on 29/Dec/21
Okay sir. Thank you so much. I  have more questions to post concerning  it.
$${Okay}\:{sir}.\:{Thank}\:{you}\:{so}\:{much}.\:{I} \\ $$$${have}\:{more}\:{questions}\:{to}\:{post}\:{concerning} \\ $$$${it}. \\ $$
Commented by mr W last updated on 29/Dec/21
assume the x−coordinate is p, then  the y−coordinate is p^2 . p is unknown.
$${assume}\:{the}\:{x}−{coordinate}\:{is}\:{p},\:{then} \\ $$$${the}\:{y}−{coordinate}\:{is}\:{p}^{\mathrm{2}} .\:{p}\:{is}\:{unknown}. \\ $$
Commented by Mathematification last updated on 30/Dec/21
Is it a rule that the point of intersection   of a parabola and a circle is always in the  form (p,  p^2 )  ?   If no, then why didn′t you assume that it  is  (p,  2p).   Sir?
$${Is}\:{it}\:{a}\:{rule}\:{that}\:{the}\:{point}\:{of}\:{intersection}\: \\ $$$${of}\:{a}\:{parabola}\:{and}\:{a}\:{circle}\:{is}\:{always}\:{in}\:{the} \\ $$$${form}\:\left({p},\:\:{p}^{\mathrm{2}} \right)\:\:?\: \\ $$$${If}\:{no},\:{then}\:{why}\:{didn}'{t}\:{you}\:{assume}\:{that}\:{it} \\ $$$${is}\:\:\left({p},\:\:\mathrm{2}{p}\right).\: \\ $$$${Sir}? \\ $$
Commented by mr W last updated on 30/Dec/21
the eqn. of the parabola is given as  y=x^2 , therefore each point on the  parabola is in the form (x, x^2 ). if x=p,  then (p, p^2 ).
$${the}\:{eqn}.\:{of}\:{the}\:{parabola}\:{is}\:{given}\:{as} \\ $$$${y}={x}^{\mathrm{2}} ,\:{therefore}\:{each}\:{point}\:{on}\:{the} \\ $$$${parabola}\:{is}\:{in}\:{the}\:{form}\:\left({x},\:{x}^{\mathrm{2}} \right).\:{if}\:{x}={p}, \\ $$$${then}\:\left({p},\:{p}^{\mathrm{2}} \right). \\ $$
Commented by Mathematification last updated on 30/Dec/21
Thank you sir
$${Thank}\:{you}\:{sir} \\ $$
Commented by Mathematification last updated on 30/Dec/21
Please sir, I need your whatsapp number,   I want to be more intimate with you.
$${Please}\:{sir},\:{I}\:{need}\:{your}\:{whatsapp}\:{number},\: \\ $$$${I}\:{want}\:{to}\:{be}\:{more}\:{intimate}\:{with}\:{you}. \\ $$
Commented by mr W last updated on 30/Dec/21
sorry sir! but i don′t want that. please  use this forum for exchange! i don′t  accept private messages.
$${sorry}\:{sir}!\:{but}\:{i}\:{don}'{t}\:{want}\:{that}.\:{please} \\ $$$${use}\:{this}\:{forum}\:{for}\:{exchange}!\:{i}\:{don}'{t} \\ $$$${accept}\:{private}\:{messages}. \\ $$
Commented by Mathematification last updated on 30/Dec/21
Okay sir
$${Okay}\:{sir} \\ $$
Commented by mr W last updated on 30/Dec/21
thanks for your understanding!
$${thanks}\:{for}\:{your}\:{understanding}! \\ $$

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