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Question-28411

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Question Number 28411 by ajfour last updated on 25/Jan/18
Commented by ajfour last updated on 25/Jan/18
The smaller sphere touches the paraboloid only at inner bottommost, point. Find the smallest value of R in terms of r, if the larger sphere touches the smaller sphere and also the walls of the paraboloid.
Thesmallerspheretouchestheparaboloidonlyatinnerbottommost,point.FindthesmallestvalueofRintermsofr,ifthelargerspheretouchesthesmallersphereandalsothewallsoftheparaboloid.
Answered by ajfour last updated on 25/Jan/18
let eq. of parabola in the plane be be y=Ax^2 . eq of circle with radius r: x^2 +(y−r)^2 =r^2 since smaller circle touches parabola at vertex only the origin is to satisfy simultaneously the eq. of smaller circle and sphere and for R to be least we are to have all four roots real namely x=0. Hrnce x^2 +(Ax^2 −r)^2 =r^2 ⇒ A^2 x^4 +(1−2Ar)x^2 =0 ⇒ x^2 (A^2 x^2 +1−2Ar)=0 ⇒ A=(1/(2r)) Eq. of bigger circle: x^2 +(y−2r−R)^2 =R^2 eq. of tangent to this circle x_1 (x−x_1 )+(y_1 −2r−R)(y−y_1 )=0 tangent to parabola ((y+y_1 )/2)=((xx_1 )/(2r)) ⇒ xx_1 −r(y+y_1 )=0 for both tangents to be same then comparing coefficient of y : 2r+R−y_1 =r ...(i) ⇒ y_1 =r+R As (x_1 , y_1 ) is on parabola, so y_1 =(x_1 ^2 /(2r)) ⇒ x_1 ^2 =2r(r+R) since x_1 ^2 +(y_1 −2r−R)^2 =R^2 2r(R+r)+r^2 =R^2 ⇒ R^( 2) −2rR−3r^2 =0 (R−r)^2 =4r^2 or R=3r .
leteq.ofparabolaintheplanebebey=Ax2.eqofcirclewithradiusr:x2+(y−r)2=r2sincesmallercircletouchesparabolaatvertexonlytheoriginistosatisfysimultaneouslytheeq.ofsmallercircleandsphereandforRtobeleastwearetohaveallfourrootsrealnamelyx=0.Hrncex2+(Ax2−r)2=r2⇒A2x4+(1−2Ar)x2=0⇒x2(A2x2+1−2Ar)=0⇒A=12rEq.ofbiggercircle:x2+(y−2r−R)2=R2eq.oftangenttothiscirclex1(x−x1)+(y1−2r−R)(y−y1)=0tangenttoparabolay+y12=xx12r⇒xx1−r(y+y1)=0forbothtangentstobesamethencomparingcoefficientofy:2r+R−y1=r…(i)⇒y1=r+RAs(x1,y1)isonparabola,soy1=x122r⇒x12=2r(r+R)sincex12+(y1−2r−R)2=R22r(R+r)+r2=R2⇒R2−2rR−3r2=0(R−r)2=4r2orR=3r.
Commented by mrW2 last updated on 25/Jan/18
I think there is following relation: the first circle r_1 =(1/(2A))=r the second circle r_2 (=R)=(3/(2A))=3r the third circle r_3 =(5/(2A))=5r .... the n−th circle r_n =((2n−1)/(2A))=(2n−1)r
Ithinkthereisfollowingrelation:thefirstcircler1=12A=rthesecondcircler2(=R)=32A=3rthethirdcircler3=52A=5r….then−thcirclern=2n−12A=(2n−1)r
Commented by ajfour last updated on 26/Jan/18
i had suspected so. Thanks Sir.
ihadsuspectedso.ThanksSir.
Answered by mrW2 last updated on 25/Jan/18
let the eqn. of parabola be y=ax^2 eqn. of circle r: x^2 +(y−r)^2 =r^2 x^2 +(ax^2 −r)^2 =r^2 a^2 x^4 −(2ar−1)x^2 =0 x^2 [a^2 x^2 −(2ar−1)]=0 x=0 x^2 =((2ar−1)/a^2 ) so that the circle touches the parabola only at one point x=0, ⇒2ar−1≤0 ⇒2ar≤1 eqn. of circle R: x^2 +(y−2r−R)^2 =R^2 x^2 +(ax^2 −2r−R)^2 =R^2 a^2 x^4 −[2a(2r+R)−1]x^2 +4r(r+R)=0 Δ=[2a(2r+R)−1]^2 −16a^2 r(r+R)=0 4a^2 (2r+R)^2 −4a(2r+R)+1−16a^2 r^2 −16a^2 rR=0 16a^2 r^2 +4a^2 R^2 +16a^2 rR−8ar−4aR+1−16a^2 r^2 −16a^2 rR=0 4(aR)^2 −4(aR)−(8ar−1)=0 aR=((4+(√(4^2 +4×4(8ar−1))))/(2×4))=((1+(√(8ar)))/2) R=((1+(√(8ar)))/(2ar))×r=((1+(√4))/1)r=3r
lettheeqn.ofparabolabey=ax2eqn.ofcircler:x2+(y−r)2=r2x2+(ax2−r)2=r2a2x4−(2ar−1)x2=0x2[a2x2−(2ar−1)]=0x=0x2=2ar−1a2sothatthecircletouchestheparabolaonlyatonepointx=0,⇒2ar−1⩽0⇒2ar⩽1eqn.ofcircleR:x2+(y−2r−R)2=R2x2+(ax2−2r−R)2=R2a2x4−[2a(2r+R)−1]x2+4r(r+R)=0Δ=[2a(2r+R)−1]2−16a2r(r+R)=04a2(2r+R)2−4a(2r+R)+1−16a2r2−16a2rR=016a2r2+4a2R2+16a2rR−8ar−4aR+1−16a2r2−16a2rR=04(aR)2−4(aR)−(8ar−1)=0aR=4+42+4×4(8ar−1)2×4=1+8ar2R=1+8ar2ar×r=1+41r=3r
Commented by ajfour last updated on 25/Jan/18
Thanks for confirming, Sir!
Thanksforconfirming,Sir!

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