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Question Number 34326 by 33 last updated on 04/May/18

find the equation of the 2D curve such that the lines (x/t) + (y/((a−t) )) = 1 are always tangent to the curve. given ′a′ is a positive real constant and ′t′ is a parameter. ( 0 < t < a )

findtheequationofthe2Dcurvesuchthatthelinesxt+y(a−t)=1arealwaystangenttothecurve.given′a′isapositiverealconstantand′t′isaparameter.(0<t<a)

Answered by MJS last updated on 05/May/18

we take two of the tangents with t_1 =p−h t_2 =p+h and search their intersection y_1 =(a−p+h)(−(1/(p−h))x+1) y_2 =(a−p−h)(−(1/(p+h))x+1) y_1 =y_2 ⇒ y_1 −y_2 =0 (a−p+h)(−(1/(p−h))x+1)−(a−p−h)(−(1/(p+h))x+1)=0 2h(−(a/(p^2 −h^2 ))x+1)=0 x=((p^2 −h^2 )/a) now remember: first idea to find the tangent in ((x),((f(x))) ) was to draw the line between (((x−h)),((f(x−h))) ) and (((x+h)),((f(x+h))) ). this is a secant. with h→0 we get the tangent. same here, just from the opposite of it: h → 0 ⇒ x=(p^2 /a); y=(((a−p)^2 )/a) p=±(√(ax)) ⇒ y=x±2(√(ax))+a

wetaketwoofthetangentswitht1=p−ht2=p+handsearchtheirintersectiony1=(a−p+h)(−1p−hx+1)y2=(a−p−h)(−1p+hx+1)y1=y2⇒y1−y2=0(a−p+h)(−1p−hx+1)−(a−p−h)(−1p+hx+1)=02h(−ap2−h2x+1)=0x=p2−h2anowremember:firstideatofindthetangentin(xf(x))wastodrawthelinebetween(x−hf(x−h))and(x+hf(x+h)).thisisasecant.withh→0wegetthetangent.samehere,justfromtheoppositeofit:h→0⇒x=p2a;y=(a−p)2ap=±ax⇒y=x±2ax+a

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