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Question Number 228499 by ajfour last updated on 18/Apr/26
Commented by ajfour last updated on 18/Apr/26
If point of contact is T(h,k) find h, as well as p.
$${If}\:{point}\:{of}\:{contact}\:{is}\:{T}\left({h},{k}\right) \\ $$$${find}\:{h},\:{as}\:{well}\:{as}\:{p}. \\ $$
Answered by nikif99 last updated on 19/Apr/26
parabola f(x)=px^2 , f ′(x)=2px cubic g(x)=a(x−1)(x−5)(x−7)= a(x^3 −13x^2 +47x−35) g(0)=a(−35)=−2⇒a=(2/(35)) g(x)=(2/(35))(x^3 −13x^2 +47x−35) g′(x)=(2/(35))(3x^2 −26x+47) Tangent of f(x) at T(h, k): y−ph^2 =2ph(x−h)⇔y=2ph_(A) x−ph^2 _(B) Tangent of g(x) at T(h, k): y−(2/(35))(h^3 −13h^2 +47h−35)=... ...=(2/(35))(3h^2 −26h+47)(x−h)⇔ y=((6h^2 −52h+94)/(35))_(A) x+((−4h^3 +26h^2 −70)/(35))_(B) A=A, B=B⇒ 2ph=((6h^2 −52h+94)/(35)) (1) −ph^2 =((−4h^3 +26h^2 −70)/(35)) (2) (1)(2)⇒ determinant (((h_1 =1.572),(p_1 =0.246),(k_1 =0.608)),((h_2 =5.933),(p_2 =−0.008),(k_2 =−0.281)),((h_3 imaginary),,))
$${parabola}\:{f}\left({x}\right)={px}^{\mathrm{2}} ,\:{f}\:'\left({x}\right)=\mathrm{2}{px} \\ $$$${cubic}\:{g}\left({x}\right)={a}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{7}\right)= \\ $$$${a}\left({x}^{\mathrm{3}} −\mathrm{13}{x}^{\mathrm{2}} +\mathrm{47}{x}−\mathrm{35}\right) \\ $$$${g}\left(\mathrm{0}\right)={a}\left(−\mathrm{35}\right)=−\mathrm{2}\Rightarrow{a}=\frac{\mathrm{2}}{\mathrm{35}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{35}}\left({x}^{\mathrm{3}} −\mathrm{13}{x}^{\mathrm{2}} +\mathrm{47}{x}−\mathrm{35}\right) \\ $$$${g}'\left({x}\right)=\frac{\mathrm{2}}{\mathrm{35}}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{26}{x}+\mathrm{47}\right) \\ $$$${Tangent}\:{of}\:{f}\left({x}\right)\:{at}\:{T}\left({h},\:{k}\right): \\ $$$${y}−{ph}^{\mathrm{2}} =\mathrm{2}{ph}\left({x}−{h}\right)\Leftrightarrow{y}=\underset{{A}} {\underbrace{\mathrm{2}{ph}}x}−\underset{{B}} {\underbrace{{ph}^{\mathrm{2}} }} \\ $$$${Tangent}\:{of}\:{g}\left({x}\right)\:{at}\:{T}\left({h},\:{k}\right): \\ $$$${y}−\frac{\mathrm{2}}{\mathrm{35}}\left({h}^{\mathrm{3}} −\mathrm{13}{h}^{\mathrm{2}} +\mathrm{47}{h}−\mathrm{35}\right)=… \\ $$$$…=\frac{\mathrm{2}}{\mathrm{35}}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{26}{h}+\mathrm{47}\right)\left({x}−{h}\right)\Leftrightarrow \\ $$$${y}=\underset{{A}} {\underbrace{\frac{\mathrm{6}{h}^{\mathrm{2}} −\mathrm{52}{h}+\mathrm{94}}{\mathrm{35}}}}{x}+\underset{{B}} {\underbrace{\frac{−\mathrm{4}{h}^{\mathrm{3}} +\mathrm{26}{h}^{\mathrm{2}} −\mathrm{70}}{\mathrm{35}}}} \\ $$$${A}={A},\:{B}={B}\Rightarrow \\ $$$$\mathrm{2}{ph}=\frac{\mathrm{6}{h}^{\mathrm{2}} −\mathrm{52}{h}+\mathrm{94}}{\mathrm{35}}\:\left(\mathrm{1}\right) \\ $$$$−{ph}^{\mathrm{2}} =\frac{−\mathrm{4}{h}^{\mathrm{3}} +\mathrm{26}{h}^{\mathrm{2}} −\mathrm{70}}{\mathrm{35}}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{2}\right)\Rightarrow \\ $$$$\begin{array}{|c|c|c|}{{h}_{\mathrm{1}} =\mathrm{1}.\mathrm{572}}&\hline{{p}_{\mathrm{1}} =\mathrm{0}.\mathrm{246}}&\hline{{k}_{\mathrm{1}} =\mathrm{0}.\mathrm{608}}\\{{h}_{\mathrm{2}} =\mathrm{5}.\mathrm{933}}&\hline{{p}_{\mathrm{2}} =−\mathrm{0}.\mathrm{008}}&\hline{{k}_{\mathrm{2}} =−\mathrm{0}.\mathrm{281}}\\{{h}_{\mathrm{3}} \:{imaginary}}&\hline{}&\hline{}\\\hline\end{array} \\ $$
Commented by nikif99 last updated on 19/Apr/26
Answered by Ghisom_ last updated on 19/Apr/26
C: y=(2/(35))x^3 −((26)/(35))x^2 +((94)/(35))x−2 P: y=px^2 ⇒ x^3 −((35p+26)/2)x^2 +47x−35=0 this must have a double solution where its derivate equals zero ⇒ x^2 −((35p+26)/3)x+((47)/3)=0 ⇒ p=((3x^2 −26x+47)/(35x)) () inserting above & transforming x^3 −47x+70=0 we know 1<x<5 ⇒ x=h=((2(√(141)))/3)sin ((arcsin ((105(√(141)))/(2209)))/3) ≈1.57201771438 ⇒ p≈.246112430362
$${C}:\:{y}=\frac{\mathrm{2}}{\mathrm{35}}{x}^{\mathrm{3}} −\frac{\mathrm{26}}{\mathrm{35}}{x}^{\mathrm{2}} +\frac{\mathrm{94}}{\mathrm{35}}{x}−\mathrm{2} \\ $$$${P}:\:{y}={px}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{3}} −\frac{\mathrm{35}{p}+\mathrm{26}}{\mathrm{2}}{x}^{\mathrm{2}} +\mathrm{47}{x}−\mathrm{35}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{must}\:\mathrm{have}\:\mathrm{a}\:\mathrm{double}\:\mathrm{solution}\:\mathrm{where} \\ $$$$\mathrm{its}\:\mathrm{derivate}\:\mathrm{equals}\:\mathrm{zero} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{35}{p}+\mathrm{26}}{\mathrm{3}}{x}+\frac{\mathrm{47}}{\mathrm{3}}=\mathrm{0}\:\Rightarrow\:{p}=\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{26}{x}+\mathrm{47}}{\mathrm{35}{x}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\right) \\ $$$$\mathrm{inserting}\:\mathrm{above}\:\&\:\mathrm{transforming} \\ $$$${x}^{\mathrm{3}} −\mathrm{47}{x}+\mathrm{70}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{know}\:\mathrm{1}<{x}<\mathrm{5}\:\Rightarrow \\ $$$${x}={h}=\frac{\mathrm{2}\sqrt{\mathrm{141}}}{\mathrm{3}}\mathrm{sin}\:\frac{\mathrm{arcsin}\:\frac{\mathrm{105}\sqrt{\mathrm{141}}}{\mathrm{2209}}}{\mathrm{3}}\:\approx\mathrm{1}.\mathrm{57201771438} \\ $$$$\Rightarrow\:{p}\approx.\mathrm{246112430362} \\ $$
Commented by ajfour last updated on 18/Apr/26
Commented by ajfour last updated on 18/Apr/26
yes and i think (2/x)=(1/(x−1))+(1/(x−5))+(1/(x−7))
$${yes} \\ $$$${and}\:{i}\:{think} \\ $$$$\frac{\mathrm{2}}{{x}}=\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{5}}+\frac{\mathrm{1}}{{x}−\mathrm{7}} \\ $$
Commented by Ghisom_ last updated on 18/Apr/26
yes, gives the same equation x^3 −47x+70=0
$$\mathrm{yes},\:\mathrm{gives}\:\mathrm{the}\:\mathrm{same}\:\mathrm{equation}\:{x}^{\mathrm{3}} −\mathrm{47}{x}+\mathrm{70}=\mathrm{0} \\ $$
Commented by Ghisom_ last updated on 18/Apr/26
...and for p<0 and 5<x<7 we get a 2^(nd) solution x=h≈5.93311089388 p≈−.00775061174689
$$…\mathrm{and}\:\mathrm{for}\:{p}<\mathrm{0}\:\mathrm{and}\:\mathrm{5}<{x}<\mathrm{7}\:\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{solution} \\ $$$${x}={h}\approx\mathrm{5}.\mathrm{93311089388} \\ $$$${p}\approx−.\mathrm{00775061174689} \\ $$
Commented by Ghisom_ last updated on 18/Apr/26
...and now I′m trying to understand why the 3^(rd) solution is false...
$$…\mathrm{and}\:\mathrm{now}\:\mathrm{I}'\mathrm{m}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{why} \\ $$$$\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{solution}\:\mathrm{is}\:\mathrm{false}… \\ $$
Commented by nikif99 last updated on 18/Apr/26
it refers to an imaginary root
$${it}\:{refers}\:{to}\:{an}\:{imaginary}\:{root} \\ $$
Commented by nikif99 last updated on 18/Apr/26
Commented by Ghisom_ last updated on 19/Apr/26
I had made a typo... of course there are 3 solutions p_1 ≈−1.56507914808 ⇒ T_1 ≈ (((−7.50512860826)),((−88.1561434127)) ) p_2 ≈.246112430362 ⇒ T_2 ≈ (((1.57201771438)),((.608202807179)) ) p_3 ≈−.00797205779680 ⇒ T_3 ≈ (((5.93311089388)),((−.280630823049)) )
$$\mathrm{I}\:\mathrm{had}\:\mathrm{made}\:\mathrm{a}\:\mathrm{typo}… \\ $$$$\mathrm{of}\:\mathrm{course}\:\mathrm{there}\:\mathrm{are}\:\mathrm{3}\:\mathrm{solutions} \\ $$$${p}_{\mathrm{1}} \approx−\mathrm{1}.\mathrm{56507914808} \\ $$$$\:\:\:\:\:\Rightarrow\:{T}_{\mathrm{1}} \approx\begin{pmatrix}{−\mathrm{7}.\mathrm{50512860826}}\\{−\mathrm{88}.\mathrm{1561434127}}\end{pmatrix} \\ $$$${p}_{\mathrm{2}} \approx.\mathrm{246112430362} \\ $$$$\:\:\:\:\:\Rightarrow\:{T}_{\mathrm{2}} \approx\begin{pmatrix}{\mathrm{1}.\mathrm{57201771438}}\\{.\mathrm{608202807179}}\end{pmatrix} \\ $$$${p}_{\mathrm{3}} \approx−.\mathrm{00797205779680} \\ $$$$\:\:\:\:\:\Rightarrow\:{T}_{\mathrm{3}} \approx\begin{pmatrix}{\mathrm{5}.\mathrm{93311089388}}\\{−.\mathrm{280630823049}}\end{pmatrix} \\ $$
Answered by mr W last updated on 19/Apr/26
g(x)=a(x−1)(x−5)(x−7) g(0)=a(0−1)(0−5)(0−7)=−2 ⇒a=(2/(35)) g(x)=(2/(35))(x^3 −13x^2 +47x−35) g′(x)=(2/(35))(3x^2 −26x+47) f(x)=px^2 f′(x)=2px (2/(35))(h^3 −13h^2 +47h−35)=ph^2 (2/(35))(3h^2 −26h+47)=2ph (2/(35))(h^3 −13h^2 +47h−35)×2=(2/(35))(3h^2 −26h+47)h ⇒h^3 −47h+70=0 this has three real roots: h_(1,2,3) =2(√((47)/3)) sin ((1/3) sin^(−1) ((105)/( 47))(√(3/(47)))+((2kπ)/3)) with k=0,1,2 (h≈1.562, 5.933, −7.505) that means there are three possible parabolas which tangent the cubic curve.
$${g}\left({x}\right)={a}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{5}\right)\left({x}−\mathrm{7}\right) \\ $$$${g}\left(\mathrm{0}\right)={a}\left(\mathrm{0}−\mathrm{1}\right)\left(\mathrm{0}−\mathrm{5}\right)\left(\mathrm{0}−\mathrm{7}\right)=−\mathrm{2} \\ $$$$\Rightarrow{a}=\frac{\mathrm{2}}{\mathrm{35}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{2}}{\mathrm{35}}\left({x}^{\mathrm{3}} −\mathrm{13}{x}^{\mathrm{2}} +\mathrm{47}{x}−\mathrm{35}\right) \\ $$$${g}'\left({x}\right)=\frac{\mathrm{2}}{\mathrm{35}}\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{26}{x}+\mathrm{47}\right) \\ $$$${f}\left({x}\right)={px}^{\mathrm{2}} \\ $$$${f}'\left({x}\right)=\mathrm{2}{px} \\ $$$$\frac{\mathrm{2}}{\mathrm{35}}\left({h}^{\mathrm{3}} −\mathrm{13}{h}^{\mathrm{2}} +\mathrm{47}{h}−\mathrm{35}\right)={ph}^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}}{\mathrm{35}}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{26}{h}+\mathrm{47}\right)=\mathrm{2}{ph} \\ $$$$\frac{\mathrm{2}}{\mathrm{35}}\left({h}^{\mathrm{3}} −\mathrm{13}{h}^{\mathrm{2}} +\mathrm{47}{h}−\mathrm{35}\right)×\mathrm{2}=\frac{\mathrm{2}}{\mathrm{35}}\left(\mathrm{3}{h}^{\mathrm{2}} −\mathrm{26}{h}+\mathrm{47}\right){h} \\ $$$$\Rightarrow{h}^{\mathrm{3}} −\mathrm{47}{h}+\mathrm{70}=\mathrm{0} \\ $$$${this}\:{has}\:{three}\:{real}\:{roots}: \\ $$$${h}_{\mathrm{1},\mathrm{2},\mathrm{3}} =\mathrm{2}\sqrt{\frac{\mathrm{47}}{\mathrm{3}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\:\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{105}}{\:\mathrm{47}}\sqrt{\frac{\mathrm{3}}{\mathrm{47}}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)\:{with}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$\left({h}\approx\mathrm{1}.\mathrm{562},\:\mathrm{5}.\mathrm{933},\:−\mathrm{7}.\mathrm{505}\right) \\ $$$${that}\:{means}\:{there}\:{are}\:{three}\:{possible} \\ $$$${parabolas}\:{which}\:{tangent}\:{the} \\ $$$${cubic}\:{curve}. \\ $$
Commented by mr W last updated on 19/Apr/26
Commented by ajfour last updated on 19/Apr/26
Elegant solution sir!
$${Elegant}\:{solution}\:{sir}!\: \\ $$

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