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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-lt-t-lt-a-

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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 )
$${find}\:{the}\:{equation}\:{of}\:{the}\:\mathrm{2}{D} \\ $$$${curve}\:{such}\:{that}\:{the}\:{lines} \\ $$$$\:\frac{{x}}{{t}}\:+\:\frac{{y}}{\left({a}−{t}\right)\:}\:=\:\mathrm{1} \\ $$$$\:{are}\:{always}\:{tangent}\:{to} \\ $$$${the}\:{curve}. \\ $$$${given}\:'{a}'\:\:{is}\:{a}\:{positive}\:{real} \\ $$$${constant}\:{and}\:'{t}'\:{is}\:{a} \\ $$$${parameter}.\:\left(\:\mathrm{0}\:<\:{t}\:<\:{a}\:\right) \\ $$
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
$$\mathrm{we}\:\mathrm{take}\:\mathrm{two}\:\mathrm{of}\:\mathrm{the}\:\mathrm{tangents}\:\mathrm{with} \\ $$$${t}_{\mathrm{1}} ={p}−{h} \\ $$$${t}_{\mathrm{2}} ={p}+{h} \\ $$$$\mathrm{and}\:\mathrm{search}\:\mathrm{their}\:\mathrm{intersection} \\ $$$${y}_{\mathrm{1}} =\left({a}−{p}+{h}\right)\left(−\frac{\mathrm{1}}{{p}−{h}}{x}+\mathrm{1}\right) \\ $$$${y}_{\mathrm{2}} =\left({a}−{p}−{h}\right)\left(−\frac{\mathrm{1}}{{p}+{h}}{x}+\mathrm{1}\right) \\ $$$${y}_{\mathrm{1}} ={y}_{\mathrm{2}} \:\Rightarrow\:{y}_{\mathrm{1}} −{y}_{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}−{p}+{h}\right)\left(−\frac{\mathrm{1}}{{p}−{h}}{x}+\mathrm{1}\right)−\left({a}−{p}−{h}\right)\left(−\frac{\mathrm{1}}{{p}+{h}}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\mathrm{2}{h}\left(−\frac{{a}}{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\frac{{p}^{\mathrm{2}} −{h}^{\mathrm{2}} }{{a}} \\ $$$$\mathrm{now}\:\mathrm{remember}: \\ $$$$\mathrm{first}\:\mathrm{idea}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{in}\:\begin{pmatrix}{{x}}\\{{f}\left({x}\right)}\end{pmatrix}\:\mathrm{was} \\ $$$$\mathrm{to}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{line}\:\mathrm{between}\:\begin{pmatrix}{{x}−{h}}\\{{f}\left({x}−{h}\right)}\end{pmatrix}\:\mathrm{and} \\ $$$$\begin{pmatrix}{{x}+{h}}\\{{f}\left({x}+{h}\right)}\end{pmatrix}.\:\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{secant}.\:\mathrm{with}\:{h}\rightarrow\mathrm{0}\:\mathrm{we} \\ $$$$\mathrm{get}\:\mathrm{the}\:\mathrm{tangent}. \\ $$$$\mathrm{same}\:\mathrm{here},\:\mathrm{just}\:\mathrm{from}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{of}\:\mathrm{it}: \\ $$$${h}\:\rightarrow\:\mathrm{0}\:\Rightarrow\:{x}=\frac{{p}^{\mathrm{2}} }{{a}};\:{y}=\frac{\left({a}−{p}\right)^{\mathrm{2}} }{{a}} \\ $$$${p}=\pm\sqrt{{ax}}\:\Rightarrow\:{y}={x}\pm\mathrm{2}\sqrt{{ax}}+{a} \\ $$

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