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Question-186835

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Question Number 186835 by ajfour last updated on 11/Feb/23
Commented by ajfour last updated on 11/Feb/23
Curve is y=x^3 −x Horizontal lines y=0 and y=c Say a root of curve be x=p. A(p,0) From A a tangent is drawn to curve; touches at Q(q, q^3 −q) say equation be y=(3q^2 −1)(x−p) q^3 −q=(3q^2 −1)(q−p) Let a normal to curve from Q and at Q intersect at S(s,s^3 −s) Let s≠q Equation of such a normal is y−(q^3 −q)=−(((x−q))/((3q^2 −1))) ⇒ s^3 −s−q^3 +q=−(((s−q))/(3q^2 −1)) ⇒ s^2 +q^2 +qs−1=(1/((1−3q^2 ))) q^3 −q=(3q^2 −1)(q−p) p^3 =p+c or s^3 −s−(3q^2 −1)(q−p)=((s−q)/(3q^2 −1)) ........................................................ while if s=q and we find where such a normal intersects the curve again, then (x^3 −x)−(q^3 −q)=−(((x−q))/((3q^2 −1))) ⇒ (x^2 +q^2 +qx−1)(3q^2 −1)+1=0 & q^3 −q=(3q^2 −1)(q−p) while p^3 =p+c ....
$${Curve}\:{is}\:{y}={x}^{\mathrm{3}} −{x} \\ $$$${Horizontal}\:{lines}\:{y}=\mathrm{0}\:{and}\:{y}={c} \\ $$$${Say}\:{a}\:{root}\:{of}\:{curve}\:{be}\:{x}={p}. \\ $$$${A}\left({p},\mathrm{0}\right)\:\:\: \\ $$$${From}\:{A}\:{a}\:{tangent}\:{is}\:{drawn}\:{to} \\ $$$${curve};\:{touches}\:{at}\:{Q}\left({q},\:{q}^{\mathrm{3}} −{q}\right) \\ $$$$\:{say}\:{equation}\:{be} \\ $$$${y}=\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right) \\ $$$${q}^{\mathrm{3}} −{q}=\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)\left({q}−{p}\right) \\ $$$${Let}\:{a}\:{normal}\:{to}\:{curve}\:{from}\:{Q}\: \\ $$$${and}\:{at}\:{Q}\:{intersect}\:{at}\:{S}\left({s},{s}^{\mathrm{3}} −{s}\right) \\ $$$${Let}\:\:{s}\neq{q} \\ $$$${Equation}\:{of}\:{such}\:{a}\:{normal}\:{is} \\ $$$${y}−\left({q}^{\mathrm{3}} −{q}\right)=−\frac{\left({x}−{q}\right)}{\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\Rightarrow\:\:{s}^{\mathrm{3}} −{s}−{q}^{\mathrm{3}} +{q}=−\frac{\left({s}−{q}\right)}{\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow\:\:{s}^{\mathrm{2}} +{q}^{\mathrm{2}} +{qs}−\mathrm{1}=\frac{\mathrm{1}}{\left(\mathrm{1}−\mathrm{3}{q}^{\mathrm{2}} \right)} \\ $$$${q}^{\mathrm{3}} −{q}=\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)\left({q}−{p}\right) \\ $$$${p}^{\mathrm{3}} ={p}+{c} \\ $$$${or} \\ $$$${s}^{\mathrm{3}} −{s}−\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)\left({q}−{p}\right)=\frac{{s}−{q}}{\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}} \\ $$$$……………………………………………….. \\ $$$${while}\:{if}\:{s}={q} \\ $$$${and}\:{we}\:{find}\:{where}\:{such}\:{a}\:{normal} \\ $$$${intersects}\:{the}\:{curve}\:{again},\:{then} \\ $$$$\left({x}^{\mathrm{3}} −{x}\right)−\left({q}^{\mathrm{3}} −{q}\right)=−\frac{\left({x}−{q}\right)}{\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$$\Rightarrow\:\left({x}^{\mathrm{2}} +{q}^{\mathrm{2}} +{qx}−\mathrm{1}\right)\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\& \\ $$$${q}^{\mathrm{3}} −{q}=\left(\mathrm{3}{q}^{\mathrm{2}} −\mathrm{1}\right)\left({q}−{p}\right) \\ $$$${while}\:\:{p}^{\mathrm{3}} ={p}+{c} \\ $$$$…. \\ $$$$ \\ $$

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