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The equation of the right Strophoid is given as $y^2(a+x)=x^2(a-x)$ and the parametric equations are: $$ x=a\sin t=a\cos u\\ y=a\tan t(1-\sin t)=a\cos u\tan\frac{u}{2} $$

Attempt

stpd

According to the definition th right Strophoid is the locus of points $P(x,y)$ for which $|O'Q|=|PQ|$

Let $\angle PON=\theta$ $$ \sin\theta=\frac{PN}{PO}=\frac{O'Q}{OQ}=\frac{MQ}{PQ}\\ \cos\theta=\frac{NO}{PO}=\frac{O'O}{OQ}=\frac{PM}{PQ}\\ \tan\theta=\frac{PN}{ON}=\frac{QM}{PM}=\frac{O'Q}{OO'}=\frac{O'Q}{a}=\frac{PQ}{a} $$ $$ x^2=PM^2=PQ^2-QM^2=PQ^2-PQ^2\sin^2\theta=PQ^2\cos^2\theta=a^2\tan^2\theta.\cos^2\theta=a^2\sin^2\theta\\ \implies\color{red}{\boxed{x=a\sin\theta}} $$ $$ \frac{PN}{QM}=\frac{PO}{PQ}\\ y^2=PN^2=\frac{PO^2}{PQ^2}.QM^2=\frac{PO^2}{a^2\tan^2\theta}.QM^2=\frac{PO^2}{a^2\tan^2\theta}.PM^2\tan^2\theta\\ =\frac{}{a^2\tan^2\theta}.a^2\sin^2\theta\tan^2\theta\\ =PO^2\sin^2\theta $$

How do I solve for $y$ here ?

Sooraj S
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    Form your diagram it is evident that $PO=QO-QP=QO-QO' $. We also have $QO=\frac{a}{\cos \theta}$ and $QO'=a\tan \theta$. Therefore, $PO=\frac{a}{\cos \theta}-a\tan \theta$. I think you can do the rest. – YNK Jan 16 '20 at 16:20

1 Answers1

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Note that \begin{align*} y^2&=PO^2-ON^2\\ &=\frac {y^2}{\sin^2\newcommand{\t}{\theta}\t}-(a-a\sin\t)^2. \end{align*} So, \begin{align*} y^2\left (\frac{\sin^2\t-1}{\sin^2\t}\right)&=-a^2 (1-\sin\t)^2. \end{align*} Now, it is easy to see that $$y=a\tan\t (1-\sin \t). $$

cqfd
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