Question-63336

Question Number 63336 by ajfour last updated on 02/Jul/19

Commented by ajfour last updated on 02/Jul/19

Find the equation of the parabola in the form: y=Ax^2 +C .

$${Find}\:{the}\:{equation}\:{of}\:{the}\:{parabola} \\ $$$${in}\:{the}\:{form}:\:\:\:{y}={Ax}^{\mathrm{2}} +{C}\:. \\ $$

Answered by mr W last updated on 07/Jul/19

y=Ax^2 +C y′=2Ax P[a(1+cos θ), a(1−sin θ)] tan θ=(1/(2Aa(1+cos θ))) let t=tan (θ/2) ((2t)/(1−t^2 ))=(1/(2Aa(1+((1−t^2 )/(1+t^2 ))))) ((2t)/(1−t^2 ))=((1+t^2 )/(4Aa)) ⇒t^4 +8Aat−1=0 ⇒8A=((1−t^4 )/(at)) a(1−sin θ)=Aa^2 (1+cos θ)^2 +C a(((1−t)^2 )/(1+t^2 ))=4Aa^2 (1/((1+t^2 )^2 ))+C ...(i) Q[b(1+cos ϕ), −b(1+sin ϕ)] tan ϕ=(1/(2Ab(1+cos ϕ))) let s=tan (ϕ/2) ⇒s^4 +8Abs−1=0 ⇒8A=((1−s^4 )/(bs)) −b(1+sin ϕ)=Ab^2 (1+cos ϕ)^2 +C −b(((1+s)^2 )/(1+s^2 ))=4Ab^2 (1/((1+s^2 )^2 ))+C ...(ii) (i)−(ii): a(((1−t)^2 )/(1+t^2 ))+b(((1+s)^2 )/(1+s^2 ))=((1−t^4 )/(2at))[(a^2 /((1+t^2 )^2 ))−(b^2 /((1+s^2 )^2 ))] ...(iii) ((1−t^4 )/(at))=((1−s^4 )/(bs)) ...(iv) ...... example: a=5, b=4 ⇒t=tan (θ/2)=0.10001 ⇒s=tan (ϕ/2)=0.12489

$${y}={Ax}^{\mathrm{2}} +{C} \\ $$$${y}'=\mathrm{2}{Ax} \\ $$$${P}\left[{a}\left(\mathrm{1}+\mathrm{cos}\:\theta\right),\:{a}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)\right] \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mathrm{2}{Aa}\left(\mathrm{1}+\mathrm{cos}\:\theta\right)} \\ $$$${let}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}{Aa}\left(\mathrm{1}+\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} }=\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{4}{Aa}} \\ $$$$\Rightarrow{t}^{\mathrm{4}} +\mathrm{8}{Aat}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{8}{A}=\frac{\mathrm{1}−{t}^{\mathrm{4}} }{{at}} \\ $$$${a}\left(\mathrm{1}−\mathrm{sin}\:\theta\right)={Aa}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} +{C} \\ $$$${a}\frac{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }=\mathrm{4}{Aa}^{\mathrm{2}} \frac{\mathrm{1}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }+{C}\:\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${Q}\left[{b}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right),\:−{b}\left(\mathrm{1}+\mathrm{sin}\:\varphi\right)\right] \\ $$$$\mathrm{tan}\:\varphi=\frac{\mathrm{1}}{\mathrm{2}{Ab}\left(\mathrm{1}+\mathrm{cos}\:\varphi\right)} \\ $$$${let}\:{s}=\mathrm{tan}\:\frac{\varphi}{\mathrm{2}} \\ $$$$\Rightarrow{s}^{\mathrm{4}} +\mathrm{8}{Abs}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{8}{A}=\frac{\mathrm{1}−{s}^{\mathrm{4}} }{{bs}} \\ $$$$ \\ $$$$−{b}\left(\mathrm{1}+\mathrm{sin}\:\varphi\right)={Ab}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{cos}\:\varphi\right)^{\mathrm{2}} +{C} \\ $$$$−{b}\frac{\left(\mathrm{1}+{s}\right)^{\mathrm{2}} }{\mathrm{1}+{s}^{\mathrm{2}} }=\mathrm{4}{Ab}^{\mathrm{2}} \frac{\mathrm{1}}{\left(\mathrm{1}+{s}^{\mathrm{2}} \right)^{\mathrm{2}} }+{C}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$${a}\frac{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }+{b}\frac{\left(\mathrm{1}+{s}\right)^{\mathrm{2}} }{\mathrm{1}+{s}^{\mathrm{2}} }=\frac{\mathrm{1}−{t}^{\mathrm{4}} }{\mathrm{2}{at}}\left[\frac{{a}^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }−\frac{{b}^{\mathrm{2}} }{\left(\mathrm{1}+{s}^{\mathrm{2}} \right)^{\mathrm{2}} }\right]\:\:\:…\left({iii}\right) \\ $$$$\frac{\mathrm{1}−{t}^{\mathrm{4}} }{{at}}=\frac{\mathrm{1}−{s}^{\mathrm{4}} }{{bs}}\:\:\:…\left({iv}\right) \\ $$$$…… \\ $$$${example}: \\ $$$${a}=\mathrm{5},\:{b}=\mathrm{4} \\ $$$$\Rightarrow{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}=\mathrm{0}.\mathrm{10001} \\ $$$$\Rightarrow{s}=\mathrm{tan}\:\frac{\varphi}{\mathrm{2}}=\mathrm{0}.\mathrm{12489} \\ $$

Commented by mr W last updated on 07/Jul/19