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

Commented by ajfour last updated on 02/Jul/19

$$Findtheequationoftheparabola$$$$intheform:y={Ax}^2+C.$$

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

$$y={Ax}^2+C$$$$y^{\prime }=2Ax$$$$P[a(1+\cos \theta ),a(1-\sin \theta )]$$$$\tan \theta =\frac{1}{2Aa(1+\cos \theta )}$$$$lett=\tan \frac{\theta }{2}$$$$\frac{2t}{1-t^2}=\frac{1}{2Aa(1+\frac{1-t^2}{1+t^2})}$$$$\frac{2t}{1-t^2}=\frac{1+t^2}{4Aa}$$$$\Rightarrow t^4+8Aat-1=0$$$$\Rightarrow 8A=\frac{1-t^4}{at}$$$$a(1-\sin \theta )={Aa}^2{(1+\cos \theta )}^2+C$$$$a\frac{{(1-t)}^2}{1+t^2}=4{Aa}^2\frac{1}{{(1+t^2)}^2}+C...\left(i\right)$$$$$$$$Q[b(1+\cos \phi ),-b(1+\sin \phi )]$$$$\tan \phi =\frac{1}{2Ab(1+\cos \phi )}$$$$lets=\tan \frac{\phi }{2}$$$$\Rightarrow s^4+8Abs-1=0$$$$\Rightarrow 8A=\frac{1-s^4}{bs}$$$$$$$$-b(1+\sin \phi )={Ab}^2{(1+\cos \phi )}^2+C$$$$-b\frac{{(1+s)}^2}{1+s^2}=4{Ab}^2\frac{1}{{(1+s^2)}^2}+C...\left(ii\right)$$$$$$$$\left(i\right)-\left(ii\right):$$$$a\frac{{(1-t)}^2}{1+t^2}+b\frac{{(1+s)}^2}{1+s^2}=\frac{1-t^4}{2at}[\frac{a^2}{{(1+t^2)}^2}-\frac{b^2}{{(1+s^2)}^2}]...\left(iii\right)$$$$\frac{1-t^4}{at}=\frac{1-s^4}{bs}...\left(iv\right)$$$$......$$$$example:$$$$a=5,b=4$$$$\Rightarrow t=\tan \frac{\theta }{2}=0.10001$$$$\Rightarrow s=\tan \frac{\phi }{2}=0.12489$$

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

Commented by mr W last updated on 08/Jul/19

$$icannotfindawaytoreducethe$$$$solutiontoasingleequation.$$

Commented by ajfour last updated on 08/Jul/19

$$Itsalrightsir,butletsinquire$$$$insomeotherfashiontoo,we$$$$mighthitluck..$$

Answered by ajfour last updated on 08/Jul/19

Commented by ajfour last updated on 08/Jul/19

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