Question-166481

Question Number 166481 by ajfour last updated on 20/Feb/22

Commented by ajfour last updated on 20/Feb/22

$${Find}\:{the}\:{equation}\:{of}\:{the}\:{parabola}. \\ $$

Answered by MJS_new last updated on 20/Feb/22

I set 2r=1 ⇒ par: y=ax^2 +(1−a) with a≈.594384753544 using the method of common tangents in P get an equation of 5^(th) degree for p=x_P p^5 −2p^4 +((25)/4)p^3 −(7/4)p^2 +(3/4)p−(1/4)=0 which I cannot solve exactly and a=((1−2p)/(4p(√(p−p^2 ))))=(1/(p^2 +1))−(p/(2(p^2 +1)=(√(p−p^2 ))))

$$\mathrm{I}\:\mathrm{set}\:\mathrm{2}{r}=\mathrm{1}\:\Rightarrow \\ $$$$\mathrm{par}:\:{y}={ax}^{\mathrm{2}} +\left(\mathrm{1}−{a}\right)\:\mathrm{with}\:{a}\approx.\mathrm{594384753544} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{common}\:\mathrm{tangents}\:\mathrm{in}\:{P} \\ $$$$\mathrm{get}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{degree}\:\mathrm{for}\:{p}={x}_{{P}} \\ $$$${p}^{\mathrm{5}} −\mathrm{2}{p}^{\mathrm{4}} +\frac{\mathrm{25}}{\mathrm{4}}{p}^{\mathrm{3}} −\frac{\mathrm{7}}{\mathrm{4}}{p}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}{p}−\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{exactly}\:\mathrm{and} \\ $$$${a}=\frac{\mathrm{1}−\mathrm{2}{p}}{\mathrm{4}{p}\sqrt{{p}−{p}^{\mathrm{2}} }}=\frac{\mathrm{1}}{{p}^{\mathrm{2}} +\mathrm{1}}−\frac{{p}}{\mathrm{2}\left({p}^{\mathrm{2}} +\mathrm{1}\right)=\sqrt{{p}−{p}^{\mathrm{2}} }} \\ $$

Answered by ajfour last updated on 21/Feb/22

Let r=1 A(1−cos θ, sin θ) ((cos θ)/(sin θ))=2a(1−cos θ) y=ax^2 +2−4a sin θ=a(1−cos θ)^2 +2−4a ⇒ a=((2−sin θ)/(4−(1−cos θ)^2 ))=((cos θ)/(2sin θ(1−cos θ))) a≈0.2972

$${Let}\:\:{r}=\mathrm{1} \\ $$$${A}\left(\mathrm{1}−\mathrm{cos}\:\theta,\:\mathrm{sin}\:\theta\right) \\ $$$$\frac{\mathrm{cos}\:\theta}{\mathrm{sin}\:\theta}=\mathrm{2}{a}\left(\mathrm{1}−\mathrm{cos}\:\theta\right) \\ $$$${y}={ax}^{\mathrm{2}} +\mathrm{2}−\mathrm{4}{a} \\ $$$$\mathrm{sin}\:\theta={a}\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{2}−\mathrm{4}{a} \\ $$$$\Rightarrow\: \\ $$$${a}=\frac{\mathrm{2}−\mathrm{sin}\:\theta}{\mathrm{4}−\left(\mathrm{1}−\mathrm{cos}\:\theta\right)^{\mathrm{2}} }=\frac{\mathrm{cos}\:\theta}{\mathrm{2sin}\:\theta\left(\mathrm{1}−\mathrm{cos}\:\theta\right)} \\ $$$${a}\approx\mathrm{0}.\mathrm{2972} \\ $$

Commented by MJS_new last updated on 21/Feb/22

yes, your r is 2 times my r ⇒ your a is half of my a

$$\mathrm{yes},\:\mathrm{your}\:{r}\:\mathrm{is}\:\mathrm{2}\:\mathrm{times}\:\mathrm{my}\:{r}\:\Rightarrow\:\mathrm{your}\:{a}\:\mathrm{is}\:\mathrm{half} \\ $$$$\mathrm{of}\:\mathrm{my}\:{a} \\ $$

Commented by ajfour last updated on 21/Feb/22

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