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Question Number 35646 by ajfour last updated on 21/May/18

Commented by ajfour last updated on 21/May/18

Find the equation of the parabola  that touches the three circles; (the  larger circle at the vertex).

$${Find}\:{the}\:{equation}\:{of}\:{the}\:{parabola} \\ $$$${that}\:{touches}\:{the}\:{three}\:{circles};\:\left({the}\right. \\ $$$$\left.{larger}\:{circle}\:{at}\:{the}\:{vertex}\right). \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 22/May/18

vertex=(0,−R)  the centre of other two circle{(R+r)sinθ,−  (R+r)cosθ} and{−(R+r)sinθ,−(R+r)cosθ}  if we can find the points where the parabola  touches the small circle then the equation of  parabola can be found  the eauation of parabola  x^2 =−4A(y+R)  this equation passes through vertex and two  cintact point  pls comment....

$${vertex}=\left(\mathrm{0},−{R}\right) \\ $$$${the}\:{centre}\:{of}\:{other}\:{two}\:{circle}\left\{\left({R}+{r}\right){sin}\theta,−\right. \\ $$$$\left.\left({R}+{r}\right){cos}\theta\right\}\:{and}\left\{−\left({R}+{r}\right){sin}\theta,−\left({R}+{r}\right){cos}\theta\right\} \\ $$$${if}\:{we}\:{can}\:{find}\:{the}\:{points}\:{where}\:{the}\:{parabola} \\ $$$${touches}\:{the}\:{small}\:{circle}\:{then}\:{the}\:{equation}\:{of} \\ $$$${parabola}\:{can}\:{be}\:{found} \\ $$$${the}\:{eauation}\:{of}\:{parabola} \\ $$$${x}^{\mathrm{2}} =−\mathrm{4}{A}\left({y}+{R}\right) \\ $$$${this}\:{equation}\:{passes}\:{through}\:{vertex}\:{and}\:{two} \\ $$$${cintact}\:{point} \\ $$$${pls}\:{comment}.... \\ $$

Answered by ajfour last updated on 22/May/18

Let required eq. of parabola be            y=Ax^2 −R  slope of common tangent to  small circle on right and parabola is  m=−(((x−(R+r)sin θ)/(y+(R+r)cos θ)))=2Ax  ..(i)  and         [y+(R+r)cos θ]^2 =r^2 −[x−(R+r)sin θ]^2                                         .......(ii)  using (ii) in (i)   4A^2 x^2 [r^2 −[x−(R+r)sin θ]^2                     = [x−(R+r)sin θ]^2   let   x−(R+r)sin θ = t  4A^2 [t+(R+r)sin θ]^2 (r^2 −t^2 )−t^2 =0                                       ........(I)          because of tangency, two    roots are real and equal while  the other two are nonreal..  let real roots be=a  while complex ones p±iq ; then    −4A^2 (t−a)^2 (t−p+iq)(t−p−iq)=0  ⇒ 4A^2 (t−a)^2 (t^2 −2pt+p^2 −q^2 )=0                                     .......(II)  comparing coefficients of  quadratic  term t^2  in (I) and (II),    4A^2 r^2 −4A^2 (R+r)^2 sin^2 θ−1            =4A^2 [a^2 +(p^2 −q^2 )+4ap]  by comparing coeff. of t^3     −8A^2 (R+r)sin θ=−8pA^2 −8aA^2   ⇒  a+p=(R+r)sin θ  by comparing coeff. of t      8A^2 r^2 (R+r)sin θ=4A^2 [−2a^2 p−2a(p^2 −q^2 )]  ........  shall have to change the method,  someone help please!

$${Let}\:{required}\:{eq}.\:{of}\:{parabola}\:{be} \\ $$$$\:\:\:\:\:\:\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{Ax}}^{\mathrm{2}} −\boldsymbol{{R}} \\ $$$${slope}\:{of}\:{common}\:{tangent}\:{to} \\ $$$${small}\:{circle}\:{on}\:{right}\:{and}\:{parabola}\:{is} \\ $$$${m}=−\left(\frac{{x}−\left({R}+{r}\right)\mathrm{sin}\:\theta}{{y}+\left({R}+{r}\right)\mathrm{cos}\:\theta}\right)=\mathrm{2}{Ax}\:\:..\left({i}\right) \\ $$$${and}\:\:\: \\ $$$$\:\:\:\:\left[{y}+\left({R}+{r}\right)\mathrm{cos}\:\theta\right]^{\mathrm{2}} ={r}^{\mathrm{2}} −\left[{x}−\left({R}+{r}\right)\mathrm{sin}\:\theta\right]^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......\left({ii}\right) \\ $$$${using}\:\left({ii}\right)\:{in}\:\left({i}\right) \\ $$$$\:\mathrm{4}{A}^{\mathrm{2}} {x}^{\mathrm{2}} \left[{r}^{\mathrm{2}} −\left[{x}−\left({R}+{r}\right)\mathrm{sin}\:\theta\right]^{\mathrm{2}} \right. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left[{x}−\left({R}+{r}\right)\mathrm{sin}\:\theta\right]^{\mathrm{2}} \\ $$$${let}\:\:\:{x}−\left({R}+{r}\right)\mathrm{sin}\:\theta\:=\:{t} \\ $$$$\mathrm{4}{A}^{\mathrm{2}} \left[{t}+\left({R}+{r}\right)\mathrm{sin}\:\theta\right]^{\mathrm{2}} \left({r}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)−{t}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:........\left({I}\right) \\ $$$$\:\:\:\:\:\:\:\:{because}\:{of}\:{tangency},\:{two} \\ $$$$\:\:{roots}\:{are}\:{real}\:{and}\:{equal}\:{while} \\ $$$${the}\:{other}\:{two}\:{are}\:{nonreal}.. \\ $$$${let}\:{real}\:{roots}\:{be}={a} \\ $$$${while}\:{complex}\:{ones}\:{p}\pm{iq}\:;\:{then} \\ $$$$\:\:−\mathrm{4}{A}^{\mathrm{2}} \left({t}−{a}\right)^{\mathrm{2}} \left({t}−{p}+{iq}\right)\left({t}−{p}−{iq}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{4}{A}^{\mathrm{2}} \left({t}−{a}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{pt}+{p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......\left({II}\right) \\ $$$${comparing}\:{coefficients}\:{of} \\ $$$${quadratic}\:\:{term}\:{t}^{\mathrm{2}} \:{in}\:\left({I}\right)\:{and}\:\left({II}\right), \\ $$$$\:\:\mathrm{4}{A}^{\mathrm{2}} {r}^{\mathrm{2}} −\mathrm{4}{A}^{\mathrm{2}} \left({R}+{r}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{4}{A}^{\mathrm{2}} \left[{a}^{\mathrm{2}} +\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)+\mathrm{4}{ap}\right] \\ $$$${by}\:{comparing}\:{coeff}.\:{of}\:{t}^{\mathrm{3}} \\ $$$$\:\:−\mathrm{8}{A}^{\mathrm{2}} \left({R}+{r}\right)\mathrm{sin}\:\theta=−\mathrm{8}{pA}^{\mathrm{2}} −\mathrm{8}{aA}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{a}+{p}=\left({R}+{r}\right)\mathrm{sin}\:\theta \\ $$$${by}\:{comparing}\:{coeff}.\:{of}\:{t} \\ $$$$\:\:\:\:\mathrm{8}{A}^{\mathrm{2}} {r}^{\mathrm{2}} \left({R}+{r}\right)\mathrm{sin}\:\theta=\mathrm{4}{A}^{\mathrm{2}} \left[−\mathrm{2}{a}^{\mathrm{2}} {p}−\mathrm{2}{a}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} \right)\right] \\ $$$$........ \\ $$$${shall}\:{have}\:{to}\:{change}\:{the}\:{method}, \\ $$$${someone}\:{help}\:{please}! \\ $$

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