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Question-72721




Question Number 72721 by mr W last updated on 01/Nov/19
Commented by mr W last updated on 01/Nov/19
find the correlation of A and B such  that the circle is inscribed between the  parabola and the x−axis as shown.
$${find}\:{the}\:{correlation}\:{of}\:{A}\:{and}\:{B}\:{such} \\ $$$${that}\:{the}\:{circle}\:{is}\:{inscribed}\:{between}\:{the} \\ $$$${parabola}\:{and}\:{the}\:{x}−{axis}\:{as}\:{shown}. \\ $$
Commented by kaivan.ahmadi last updated on 01/Nov/19
(d,2R) ∈ y=Ax+Bx^2 ⇒  2R=Ad+Bd^2 ⇒A=((2R−Bd^2 )/d)
$$\left({d},\mathrm{2}{R}\right)\:\in\:{y}={Ax}+{Bx}^{\mathrm{2}} \Rightarrow \\ $$$$\mathrm{2}{R}={Ad}+{Bd}^{\mathrm{2}} \Rightarrow{A}=\frac{\mathrm{2}{R}−{Bd}^{\mathrm{2}} }{{d}} \\ $$$$ \\ $$
Commented by mr W last updated on 01/Nov/19
not correct sir!   the parabola should just tangent the   circle.
$${not}\:{correct}\:{sir}!\: \\ $$$${the}\:{parabola}\:{should}\:{just}\:{tangent}\:{the}\: \\ $$$${circle}. \\ $$
Commented by mr W last updated on 01/Nov/19
Commented by kaivan.ahmadi last updated on 01/Nov/19
in this case first we hava (x−d)^2 +(y−R)^2 =R^2   and y=Ax+Bx^2  .  we must find the roots of equation  (x−d)^2 +(Ax+Bx^2 −R)^2 =R^2
$${in}\:{this}\:{case}\:{first}\:{we}\:{hava}\:\left({x}−{d}\right)^{\mathrm{2}} +\left({y}−{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${and}\:{y}={Ax}+{Bx}^{\mathrm{2}} \:. \\ $$$${we}\:{must}\:{find}\:{the}\:{roots}\:{of}\:{equation} \\ $$$$\left({x}−{d}\right)^{\mathrm{2}} +\left({Ax}+{Bx}^{\mathrm{2}} −{R}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$$ \\ $$
Commented by ajfour last updated on 01/Nov/19
Commented by ajfour last updated on 02/Nov/19
y=ax^2   (x−h)^2 +(y−k)^2 =r^2   ⇒  (x−h)^2 +(ax^2 −k)^2 =r^2   with  2ax=−((x−h)/(ax^2 −k))  ⇒  2a^2 x^3 +(1−2ak)x−h=0      x^3 +(((1−2ak)/(2a^2 )))x−(h/(2a^2 ))=0   D<0      the inevitable real root from  Cardano′s solution be x_0 .  say  x_0 =f(a, k)  ⇒  ax_0 ^2 =k−(√(r^2 −(x_0 −h)^2 ))  with  k+r=a(h+d)^2   _________________________  Y=AX+BX^2   X_0 =−(A/(2B))=h+d  Y_(max) =−(A^2 /(4B))=a(h+d)^2 =k+r  hence A and B corelated,  after proper substitutions.
$${y}={ax}^{\mathrm{2}} \\ $$$$\left({x}−{h}\right)^{\mathrm{2}} +\left({y}−{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left({x}−{h}\right)^{\mathrm{2}} +\left({ax}^{\mathrm{2}} −{k}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$${with}\:\:\mathrm{2}{ax}=−\frac{{x}−{h}}{{ax}^{\mathrm{2}} −{k}} \\ $$$$\Rightarrow\:\:\mathrm{2}{a}^{\mathrm{2}} {x}^{\mathrm{3}} +\left(\mathrm{1}−\mathrm{2}{ak}\right){x}−{h}=\mathrm{0} \\ $$$$\:\:\:\:{x}^{\mathrm{3}} +\left(\frac{\mathrm{1}−\mathrm{2}{ak}}{\mathrm{2}{a}^{\mathrm{2}} }\right){x}−\frac{{h}}{\mathrm{2}{a}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\:{D}<\mathrm{0}\:\:\: \\ $$$$\:{the}\:{inevitable}\:{real}\:{root}\:{from} \\ $$$${Cardano}'{s}\:{solution}\:{be}\:{x}_{\mathrm{0}} . \\ $$$${say}\:\:{x}_{\mathrm{0}} ={f}\left({a},\:{k}\right) \\ $$$$\Rightarrow\:\:{ax}_{\mathrm{0}} ^{\mathrm{2}} ={k}−\sqrt{{r}^{\mathrm{2}} −\left({x}_{\mathrm{0}} −{h}\right)^{\mathrm{2}} } \\ $$$${with}\:\:{k}+{r}={a}\left({h}+{d}\right)^{\mathrm{2}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${Y}={AX}+{BX}^{\mathrm{2}} \\ $$$${X}_{\mathrm{0}} =−\frac{{A}}{\mathrm{2}{B}}={h}+{d} \\ $$$${Y}_{{max}} =−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}}={a}\left({h}+{d}\right)^{\mathrm{2}} ={k}+{r} \\ $$$${hence}\:{A}\:{and}\:{B}\:{corelated}, \\ $$$${after}\:{proper}\:{substitutions}. \\ $$
Commented by mr W last updated on 02/Nov/19
thanks sir!  we only ensure that parabola tangents  the circle, but both can intersect at  an other point. we should find the  condition that both intersect at one  single point or one double point.
$${thanks}\:{sir}! \\ $$$${we}\:{only}\:{ensure}\:{that}\:{parabola}\:{tangents} \\ $$$${the}\:{circle},\:{but}\:{both}\:{can}\:{intersect}\:{at} \\ $$$${an}\:{other}\:{point}.\:{we}\:{should}\:{find}\:{the} \\ $$$${condition}\:{that}\:{both}\:{intersect}\:{at}\:{one} \\ $$$${single}\:{point}\:{or}\:{one}\:{double}\:{point}. \\ $$
Commented by MJS last updated on 02/Nov/19
Commented by MJS last updated on 02/Nov/19
I once posted this, it might help here. But I'm afraid we cannot generally solve.
Commented by ajfour last updated on 02/Nov/19
thanks for sharing it again, mjs sir, but herein by simply imposing D<0 would suffice for the matter is that of a cubic and not quartic, i guess.
Commented by mr W last updated on 02/Nov/19
thank you for this material sir!
$${thank}\:{you}\:{for}\:{this}\:{material}\:{sir}! \\ $$
Answered by mr W last updated on 02/Nov/19
Commented by mr W last updated on 02/Nov/19
we may have two cases:  case 1: parabola touches the circle at  two points  case 2: parabola touches the circle only  at one point    case 1:  y=Ax+Bx^2 =B[x^2 +(A/B)x+((A/(2B)))^2 ]−(A^2 /(4B))  y=B(x+(A/(2B)))^2 −(A^2 /(4B))  ⇒(A/(2B))=−d  y=B(x−d)^2 −Bd^2 =B(x^2 −2dx)  y′=2B(x−d)  point B(d−R sin ϕ, R(1+cos ϕ))  tan ϕ=2B(d−R sin ϕ−d)  ⇒B=−(1/(2R cos ϕ))  R(1+cos ϕ)=B(d−R sin ϕ−2d)(d−R sin ϕ)  R(1+cos ϕ)=((d^2 −R^2  sin^2  ϕ)/(2R cos ϕ))  1+cos ϕ=((δ^2 −sin^2  ϕ)/(2 cos ϕ)) with δ=(d/R)  (cos ϕ+1)^2 =δ^2   ⇒cos ϕ=δ−1   ⇒B=−(1/(2R(δ−1)))  i.e. we have case 1 if 1<δ≤2.   in this case the eqn. of parabola is  y=−((x(x−2d))/(2R(δ−1)))  or A=−(δ/(δ−1)), B=−(1/(2R(δ−1)))    case 2:  y=Ax+Bx^2   y′=A+2Bx  point B(d−R sin ϕ, R(1+cos ϕ))  tan ϕ=A+2B(d−R sin ϕ)   ...(i)  R(1+cos ϕ)=(d−R sin ϕ)[A+B(d−R sin ϕ)]  ((R(1+cos ϕ))/((d−R sin ϕ)))=A+B(d−R sin ϕ)  ...(ii)  (i)−(ii):  ⇒B′=BR=(((δ−sin ϕ)tan ϕ−(1+cos ϕ))/((δ−sin ϕ)^2 ))  2(ii)−(i):  ⇒A=((2(1+cos ϕ)−(δ−sin ϕ)tan ϕ)/(δ−sin ϕ))
$${we}\:{may}\:{have}\:{two}\:{cases}: \\ $$$${case}\:\mathrm{1}:\:{parabola}\:{touches}\:{the}\:{circle}\:{at} \\ $$$${two}\:{points} \\ $$$${case}\:\mathrm{2}:\:{parabola}\:{touches}\:{the}\:{circle}\:{only} \\ $$$${at}\:{one}\:{point} \\ $$$$ \\ $$$$\boldsymbol{{case}}\:\mathrm{1}: \\ $$$${y}={Ax}+{Bx}^{\mathrm{2}} ={B}\left[{x}^{\mathrm{2}} +\frac{{A}}{{B}}{x}+\left(\frac{{A}}{\mathrm{2}{B}}\right)^{\mathrm{2}} \right]−\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}} \\ $$$${y}={B}\left({x}+\frac{{A}}{\mathrm{2}{B}}\right)^{\mathrm{2}} −\frac{{A}^{\mathrm{2}} }{\mathrm{4}{B}} \\ $$$$\Rightarrow\frac{{A}}{\mathrm{2}{B}}=−{d} \\ $$$${y}={B}\left({x}−{d}\right)^{\mathrm{2}} −{Bd}^{\mathrm{2}} ={B}\left({x}^{\mathrm{2}} −\mathrm{2}{dx}\right) \\ $$$${y}'=\mathrm{2}{B}\left({x}−{d}\right) \\ $$$${point}\:{B}\left({d}−{R}\:\mathrm{sin}\:\varphi,\:{R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)\right) \\ $$$$\mathrm{tan}\:\varphi=\mathrm{2}{B}\left({d}−{R}\:\mathrm{sin}\:\varphi−{d}\right) \\ $$$$\Rightarrow{B}=−\frac{\mathrm{1}}{\mathrm{2}{R}\:\mathrm{cos}\:\varphi} \\ $$$${R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)={B}\left({d}−{R}\:\mathrm{sin}\:\varphi−\mathrm{2}{d}\right)\left({d}−{R}\:\mathrm{sin}\:\varphi\right) \\ $$$${R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)=\frac{{d}^{\mathrm{2}} −{R}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\varphi}{\mathrm{2}{R}\:\mathrm{cos}\:\varphi} \\ $$$$\mathrm{1}+\mathrm{cos}\:\varphi=\frac{\delta^{\mathrm{2}} −\mathrm{sin}^{\mathrm{2}} \:\varphi}{\mathrm{2}\:\mathrm{cos}\:\varphi}\:{with}\:\delta=\frac{{d}}{{R}} \\ $$$$\left(\mathrm{cos}\:\varphi+\mathrm{1}\right)^{\mathrm{2}} =\delta^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\varphi=\delta−\mathrm{1}\: \\ $$$$\Rightarrow{B}=−\frac{\mathrm{1}}{\mathrm{2}{R}\left(\delta−\mathrm{1}\right)} \\ $$$${i}.{e}.\:{we}\:{have}\:{case}\:\mathrm{1}\:{if}\:\mathrm{1}<\delta\leqslant\mathrm{2}.\: \\ $$$${in}\:{this}\:{case}\:{the}\:{eqn}.\:{of}\:{parabola}\:{is} \\ $$$${y}=−\frac{{x}\left({x}−\mathrm{2}{d}\right)}{\mathrm{2}{R}\left(\delta−\mathrm{1}\right)} \\ $$$${or}\:{A}=−\frac{\delta}{\delta−\mathrm{1}},\:{B}=−\frac{\mathrm{1}}{\mathrm{2}{R}\left(\delta−\mathrm{1}\right)} \\ $$$$ \\ $$$$\boldsymbol{{case}}\:\mathrm{2}: \\ $$$${y}={Ax}+{Bx}^{\mathrm{2}} \\ $$$${y}'={A}+\mathrm{2}{Bx} \\ $$$${point}\:{B}\left({d}−{R}\:\mathrm{sin}\:\varphi,\:{R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)\right) \\ $$$$\mathrm{tan}\:\varphi={A}+\mathrm{2}{B}\left({d}−{R}\:\mathrm{sin}\:\varphi\right)\:\:\:…\left({i}\right) \\ $$$${R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)=\left({d}−{R}\:\mathrm{sin}\:\varphi\right)\left[{A}+{B}\left({d}−{R}\:\mathrm{sin}\:\varphi\right)\right] \\ $$$$\frac{{R}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)}{\left({d}−{R}\:\mathrm{sin}\:\varphi\right)}={A}+{B}\left({d}−{R}\:\mathrm{sin}\:\varphi\right)\:\:…\left({ii}\right) \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\Rightarrow{B}'={BR}=\frac{\left(\delta−\mathrm{sin}\:\varphi\right)\mathrm{tan}\:\varphi−\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)}{\left(\delta−\mathrm{sin}\:\varphi\right)^{\mathrm{2}} } \\ $$$$\mathrm{2}\left({ii}\right)−\left({i}\right): \\ $$$$\Rightarrow{A}=\frac{\mathrm{2}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)−\left(\delta−\mathrm{sin}\:\varphi\right)\mathrm{tan}\:\varphi}{\delta−\mathrm{sin}\:\varphi} \\ $$
Commented by ajfour last updated on 02/Nov/19
the second case does get   parametric, Sir. I tried but  dint post.
$${the}\:{second}\:{case}\:{does}\:{get}\: \\ $$$${parametric},\:{Sir}.\:{I}\:{tried}\:{but} \\ $$$${dint}\:{post}. \\ $$
Commented by mr W last updated on 02/Nov/19
but with my method things can occur  like this:
$${but}\:{with}\:{my}\:{method}\:{things}\:{can}\:{occur} \\ $$$${like}\:{this}: \\ $$
Commented by mr W last updated on 02/Nov/19
Commented by ajfour last updated on 02/Nov/19
Keen observation Sir, i just  posted, please view..
$${Keen}\:{observation}\:{Sir},\:{i}\:{just} \\ $$$${posted},\:{please}\:{view}.. \\ $$

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