Menu Close

The-line-y-mx-meets-the-parabola-y-x-a-b-x-tangentially-where-0-lt-a-lt-b-Show-that-m-b-a-2-

Tinku Tara 0 Comments
Pin




Question Number 96244 by Don08q last updated on 31/May/20
The line y = mx meets the parabola y = (x − a)(b − x) tangentially where 0 < a < b. Show that m = ((√b) − (√a))^2
$$ \\ $$$$\:\:\mathrm{The}\:\mathrm{line}\:{y}\:=\:{mx}\:\:\mathrm{meets}\:\mathrm{the}\:\mathrm{parabola} \\ $$$$\:\:{y}\:=\:\left({x}\:−\:{a}\right)\left({b}\:−\:{x}\right)\:\mathrm{tangentially}\:\mathrm{where} \\ $$$$\:\:\mathrm{0}\:<\:{a}\:<\:{b}.\:\mathrm{Show}\:\mathrm{that}\:{m}\:=\:\left(\sqrt{{b}}\:−\:\sqrt{{a}}\right)^{\mathrm{2}} \\ $$$$ \\ $$
Commented by bobhans last updated on 31/May/20
y = −x^2 +bx+ax−ab y = −x^2 +(a+b)x−ab with slope = −2x+a+b = m x = ((a+b−m)/2) ⇒ −x^2 +(a+b)x−ab=mx x^2 +(m−a−b)x+ab = 0 set a+b = p ⇒(((p−m)/2))^2 +(m−p)(((p−m)/2))+ab = 0 −(p−m)^2 = −4ab ⇒p−m = ± 2(√(ab)) m = p ±2(√(ab)) = (a+b±2(√(ab))) m = ((√a) ± (√b) )^2
$$\mathrm{y}\:=\:−{x}^{\mathrm{2}} +{bx}+{ax}−{ab}\: \\ $$$${y}\:=\:−{x}^{\mathrm{2}} +\left({a}+{b}\right){x}−{ab}\: \\ $$$${with}\:{slope}\:=\:−\mathrm{2}{x}+{a}+{b}\:=\:{m}\: \\ $$$${x}\:=\:\frac{{a}+{b}−{m}}{\mathrm{2}}\:\Rightarrow\:−{x}^{\mathrm{2}} +\left({a}+{b}\right){x}−{ab}={mx} \\ $$$${x}^{\mathrm{2}} +\left({m}−{a}−{b}\right){x}+{ab}\:=\:\mathrm{0} \\ $$$$\mathrm{set}\:{a}+{b}\:=\:{p} \\ $$$$\Rightarrow\left(\frac{{p}−{m}}{\mathrm{2}}\right)^{\mathrm{2}} +\left({m}−{p}\right)\left(\frac{{p}−{m}}{\mathrm{2}}\right)+{ab}\:=\:\mathrm{0} \\ $$$$−\left({p}−{m}\right)^{\mathrm{2}} \:=\:−\mathrm{4}{ab}\:\Rightarrow{p}−{m}\:=\:\pm\:\mathrm{2}\sqrt{{ab}} \\ $$$${m}\:=\:{p}\:\pm\mathrm{2}\sqrt{{ab}}\:=\:\left({a}+{b}\pm\mathrm{2}\sqrt{{ab}}\right) \\ $$$${m}\:=\:\left(\sqrt{{a}}\:\pm\:\sqrt{{b}}\:\right)^{\mathrm{2}} \: \\ $$
Commented by Don08q last updated on 31/May/20
Thank you Sir.
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$
Answered by Smail last updated on 31/May/20
mx=(x−a)(b−x) at the intersection mx=−x^2 +(b+a)x−ab x^2 −(a+b−m)x+ab=0 Since the line is tangent to the parabola : Δ=0=(a+b−m)^2 −4ab a+b−m=+_− 2(√(ab)) a+b+_− 2(√(ab))=m m=((√a)+_− (√b))^2 Thus, there are two passible lines whose slopes are m_1 =((√a)−(√b))^2 and m_2 =((√a)+(√b))^2
$${mx}=\left({x}−{a}\right)\left({b}−{x}\right)\:{at}\:{the}\:{intersection} \\ $$$${mx}=−{x}^{\mathrm{2}} +\left({b}+{a}\right){x}−{ab}\: \\ $$$${x}^{\mathrm{2}} −\left({a}+{b}−{m}\right){x}+{ab}=\mathrm{0} \\ $$$${Since}\:{the}\:{line}\:{is}\:{tangent}\: \\ $$$${to}\:{the}\:{parabola}\::\:\Delta=\mathrm{0}=\left({a}+{b}−{m}\right)^{\mathrm{2}} −\mathrm{4}{ab} \\ $$$${a}+{b}−{m}=\underset{−} {+}\mathrm{2}\sqrt{{ab}} \\ $$$${a}+{b}\underset{−} {+}\mathrm{2}\sqrt{{ab}}={m} \\ $$$${m}=\left(\sqrt{{a}}\underset{−} {+}\sqrt{{b}}\right)^{\mathrm{2}} \\ $$$${Thus},\:{there}\:{are}\:{two}\:{passible} \\ $$$${lines}\:{whose}\:{slopes}\:{are}\:{m}_{\mathrm{1}} =\left(\sqrt{{a}}−\sqrt{{b}}\right)^{\mathrm{2}} \\ $$$${and}\:{m}_{\mathrm{2}} =\left(\sqrt{{a}}+\sqrt{{b}}\right)^{\mathrm{2}} \\ $$
Commented by Don08q last updated on 31/May/20
Thank you Sir.
$$\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *