Question-72721

Question Number 72721 by mr W last updated on 01/Nov/19

Commented by mr W last updated on 01/Nov/19

$${find}\:{the}\:{correlation}\:{of}\:{A}\:{and}\:{B}\:{such} \\ $$$${that}\:{the}\:{circle}\:{is}\:{inscribed}\:{between}\:{the} \\ $$$${parabola}\:{and}\:{the}\:{x}−{axis}\:{as}\:{shown}. \\ $$

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

$${not}\:{correct}\:{sir}!\: \\ $$$${the}\:{parabola}\:{should}\:{just}\:{tangent}\:{the}\: \\ $$$${circle}. \\ $$

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

$${thanks}\:{sir}! \\ $$$${we}\:{only}\:{ensure}\:{that}\:{parabola}\:{tangents} \\ $$$${the}\:{circle},\:{but}\:{both}\:{can}\:{intersect}\:{at} \\ $$$${an}\:{other}\:{point}.\:{we}\:{should}\:{find}\:{the} \\ $$$${condition}\:{that}\:{both}\:{intersect}\:{at}\:{one} \\ $$$${single}\:{point}\:{or}\:{one}\:{double}\:{point}. \\ $$

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

$${thank}\:{you}\:{for}\:{this}\:{material}\:{sir}! \\ $$

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