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Question Number 213888 by mnjuly1970 last updated on 20/Nov/24
Commented by mnjuly1970 last updated on 20/Nov/24
circle is tangant to parabola .A is center of circle Find , inf ( R )=?
$$\:\:\:\:\:{circle}\:{is}\:{tangant}\:{to}\:{parabola} \\ $$$$\:\:.{A}\:{is}\:{center}\:{of}\:{circle} \\ $$$$\:\:\:\:{Find}\:\:\:\:,\:\:{inf}\:\left(\:{R}\:\right)=? \\ $$
Commented by mr W last updated on 21/Nov/24
y=x^r ? A(0, α) ? inf(R) means what?
$${y}={x}^{{r}} \:\:\:\:? \\ $$$${A}\left(\mathrm{0},\:\alpha\right)\:\:\:\:? \\ $$$${inf}\left({R}\right)\:{means}\:{what}? \\ $$
Commented by mehdee7396 last updated on 21/Nov/24
y=x^2 A(0,a)
$${y}={x}^{\mathrm{2}} \\ $$$${A}\left(\mathrm{0},{a}\right)\: \\ $$
Commented by mnjuly1970 last updated on 21/Nov/24
infimum inf {x ∣ x∈[0 ,1)}=inf{x∣x∈(0,1)}=1
$$\:\:\:{infimum}\: \\ $$$$\:\:\:{inf}\:\left\{{x}\:\mid\:{x}\in\left[\mathrm{0}\:,\mathrm{1}\right)\right\}={inf}\left\{{x}\mid{x}\in\left(\mathrm{0},\mathrm{1}\right)\right\}=\mathrm{1}\:\: \\ $$
Commented by mr W last updated on 21/Nov/24
i don′t understand the sense to ask for inf(R) here.
$${i}\:{don}'{t}\:{understand}\:{the}\:{sense}\:{to}\:{ask} \\ $$$${for}\:{inf}\left({R}\right)\:{here}. \\ $$
Answered by mr W last updated on 21/Nov/24
y=x^2 x^2 +(y−a)^2 =R^2 y+(y−a)^2 =R^2 y^2 −(2a−1)y+a^2 −R^2 =0 Δ=(2a−1)^2 −4(a^2 −R^2 )=0 4R^2 =4a−1 ⇒R=(√(a−(1/4))) a≥R=(√(a−(1/4))) a^2 −a+(1/4)≥0 ⇒a≥(1/2)
$${y}={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${y}+\left({y}−{a}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${y}^{\mathrm{2}} −\left(\mathrm{2}{a}−\mathrm{1}\right){y}+{a}^{\mathrm{2}} −{R}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\left(\mathrm{2}{a}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left({a}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{4}{R}^{\mathrm{2}} =\mathrm{4}{a}−\mathrm{1} \\ $$$$\Rightarrow{R}=\sqrt{{a}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${a}\geqslant{R}=\sqrt{{a}−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${a}^{\mathrm{2}} −{a}+\frac{\mathrm{1}}{\mathrm{4}}\geqslant\mathrm{0}\: \\ $$$$\Rightarrow{a}\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 21/Nov/24
Commented by mnjuly1970 last updated on 21/Nov/24
thanks alot : inf(R)= 1/2
$$\:\:{thanks}\:{alot}\::\:\:{inf}\left({R}\right)=\:\mathrm{1}/\mathrm{2} \\ $$
Commented by mr W last updated on 21/Nov/24
Answered by BHOOPENDRA last updated on 21/Nov/24
Equation of circle which center A(0,a) x^2 +(y−a)^2 =R^2 tangency y=x^2 So x^2 +(x^2 −a)^2 =R^2 put x^2 =y For the circle to be tangent to the parabola the qudratic must have a repeated root.The discriminat of qudratic must be zero. Δ=b^2 −4ac (2a−1)^2 =4(a^2 −R^2 ) R^2 =(a−(1/4)) The question ask for the smallest possible value of the radius R of a circle centered at A(0,a) that touches the parabola y=x^2 at exactly two points. The infimum of R is smallest valueR can approach (but not necessrily reach).
$${Equation}\:{of}\:{circle} \\ $$$${which}\:{center}\:{A}\left(\mathrm{0},{a}\right) \\ $$$${x}^{\mathrm{2}} +\left({y}−{a}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${tangency}\: \\ $$$${y}={x}^{\mathrm{2}} \:{So} \\ $$$${x}^{\mathrm{2}} +\left({x}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} \\ $$$${put}\:{x}^{\mathrm{2}} ={y} \\ $$$${For}\:{the}\:{circle}\:{to}\:{be}\:{tangent}\:{to}\:{the}\:{parabola} \\ $$$${the}\:{qudratic}\:{must}\:{have}\:{a}\:{repeated} \\ $$$${root}.{The}\:{discriminat}\:{of}\:{qudratic} \\ $$$${must}\:{be}\:{zero}. \\ $$$$\Delta={b}^{\mathrm{2}} −\mathrm{4}{ac} \\ $$$$\:\left(\mathrm{2}{a}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left({a}^{\mathrm{2}} −{R}^{\mathrm{2}} \right) \\ $$$${R}^{\mathrm{2}} =\left({a}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${The}\:{question}\:{ask}\:{for}\:{the}\:{smallest} \\ $$$${possible}\:{value}\:{of}\:{the}\:{radius}\:{R}\:{of}\: \\ $$$${a}\:{circle}\:{centered}\:{at}\:{A}\left(\mathrm{0},{a}\right)\:{that}\: \\ $$$${touches}\:{the}\:{parabola}\:{y}={x}^{\mathrm{2}} \:{at}\:{exactly} \\ $$$${two}\:{points}. \\ $$$${The}\:{infimum}\:{of}\:{R}\:{is}\:{smallest}\:{valueR} \\ $$$${can}\:{approach}\:\left({but}\:{not}\:{necessrily}\:{reach}\right). \\ $$$$ \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 21/Nov/24
$$\:\cancel{\underline{ }} \\ $$
Commented by BHOOPENDRA last updated on 21/Nov/24
Key observation for R>0 ,the condition is:a>(1/4) The radius R minimized when a approaches its lower bound,a=(1/4). At this value R^2 =1/4 −1/4=0 While the infimum of R^2 mathematically 0 ,the smallest physical radiusR is achieved just above a=(1/4) ,where R start increasing from 0. Infimum of the radius R is 0 But it will not satisfy tangential condition So R^2 =a−1/4 put in the equation y^2 +(2a−1)y+(a^2 −R^2 )=0 y^2 +(2a−1)y+(a^2 −a+(1/4))=0 y^2 +(2a−1)y+(a−(1/2))^2 =0 a−(1/2)≥0 a≥(1/2) R≥(1/2) Inf(R)=1/2 (according to tangential condition)
$${Key}\:{observation} \\ $$$${for}\:{R}>\mathrm{0}\:,{the}\:{condition}\:{is}:{a}>\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${The}\:{radius}\:{R}\:{minimized}\:{when}\:{a} \\ $$$${approaches}\:{its}\:{lower}\:{bound},{a}=\frac{\mathrm{1}}{\mathrm{4}}. \\ $$$${At}\:{this}\:{value} \\ $$$${R}^{\mathrm{2}} =\mathrm{1}/\mathrm{4}\:−\mathrm{1}/\mathrm{4}=\mathrm{0} \\ $$$${While}\:{the}\:{infimum}\:{of}\:{R}^{\mathrm{2}} \:{mathematically}\:\mathrm{0} \\ $$$$,{the}\:{smallest}\:{physical}\:{radiusR}\:{is} \\ $$$${achieved}\:{just}\:{above}\:{a}=\frac{\mathrm{1}}{\mathrm{4}}\:,{where} \\ $$$${R}\:{start}\:{increasing}\:{from}\:\mathrm{0}. \\ $$$$ \\ $$$${Infimum}\:{of}\:{the}\:{radius}\:{R}\:{is}\:\mathrm{0} \\ $$$${But}\:{it}\:{will}\:{not}\:{satisfy}\:{tangential} \\ $$$${condition}\: \\ $$$${So}\:{R}^{\mathrm{2}} ={a}−\mathrm{1}/\mathrm{4}\:\:{put}\:{in}\:{the}\:{equation} \\ $$$${y}^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{1}\right){y}+\left({a}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{1}\right){y}+\left({a}^{\mathrm{2}} −{a}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$${y}^{\mathrm{2}} +\left(\mathrm{2}{a}−\mathrm{1}\right){y}+\left({a}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${a}−\frac{\mathrm{1}}{\mathrm{2}}\geqslant\mathrm{0} \\ $$$${a}\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${Inf}\left({R}\right)=\mathrm{1}/\mathrm{2}\:\left({according}\:{to}\:{tangential}\right. \\ $$$$\left.{condition}\right) \\ $$
Commented by mr W last updated on 21/Nov/24
a circle with radius 0 is not a circle, but just a point. such that the circle tangents the parabolla as requested, a≥(1/2) and correspondingly R≥(1/2).
$${a}\:{circle}\:{with}\:{radius}\:\mathrm{0}\:{is}\:{not}\:{a}\:{circle}, \\ $$$${but}\:{just}\:{a}\:{point}.\:{such}\:{that}\:{the}\:{circle} \\ $$$${tangents}\:{the}\:{parabolla}\:{as}\:{requested}, \\ $$$${a}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\:{and}\:{correspondingly}\:{R}\geqslant\frac{\mathrm{1}}{\mathrm{2}}. \\ $$
Commented by BHOOPENDRA last updated on 21/Nov/24
I did not say this Sir W i was just explaning inf(R) .i alredy said this R can never be zero
$${I}\:{did}\:{not}\:{say}\:{this}\:{Sir}\:{W}\:{i}\:{was}\:{just}\:{explaning} \\ $$$${inf}\left({R}\right)\:.{i}\:{alredy}\:{said}\:{this}\:{R}\:{can}\:{never} \\ $$$${be}\:{zero} \\ $$
Commented by mr W last updated on 21/Nov/24
in current case “a>(1/4) and R>0” is not true, it should be “a≥(1/2) and R≥(1/2)”.
$${in}\:{current}\:{case}\:“{a}>\frac{\mathrm{1}}{\mathrm{4}}\:{and}\:{R}>\mathrm{0}'' \\ $$$${is}\:{not}\:{true},\:{it}\:{should}\:{be}\:“{a}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\:{and} \\ $$$${R}\geqslant\frac{\mathrm{1}}{\mathrm{2}}''. \\ $$
Commented by BHOOPENDRA last updated on 21/Nov/24
whatever you got its correct but In the context of intersection point y^2 −(2a−1)y+(a^2 −R^2 )=0 substitute R^2 =a−1/4 y^2 −(2a−1)y+(a^2 −a+(1/4))=0 y^2 −(2a−1)y+(a−(1/2))^2 =0 for this qudratic to have real solution a−(1/2), ⇒a≥(1/2) R≥(1/2) i know this i was just explaning inf(R). BTW thanks
$${whatever}\:{you}\:{got}\:{its}\:{correct}\:{but} \\ $$$${In}\:{the}\:{context}\:{of}\:{intersection}\:{point} \\ $$$${y}^{\mathrm{2}} −\left(\mathrm{2}{a}−\mathrm{1}\right){y}+\left({a}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${substitute}\:{R}^{\mathrm{2}} ={a}−\mathrm{1}/\mathrm{4} \\ $$$${y}^{\mathrm{2}} −\left(\mathrm{2}{a}−\mathrm{1}\right){y}+\left({a}^{\mathrm{2}} −{a}+\frac{\mathrm{1}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$${y}^{\mathrm{2}} −\left(\mathrm{2}{a}−\mathrm{1}\right){y}+\left({a}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${for}\:{this}\:{qudratic}\:{to}\:{have}\:{real}\:{solution} \\ $$$${a}−\frac{\mathrm{1}}{\mathrm{2}}, \\ $$$$\Rightarrow{a}\geqslant\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${R}\geqslant\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$${i}\:{know}\:{this}\: \\ $$$${i}\:{was}\:{just}\:{explaning}\:{inf}\left({R}\right). \\ $$$${BTW}\:{thanks}\: \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 21/Nov/24
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