Question

If $ST$ and $SN$ are the lengths of the subtangent and the subnormal at the point $θ=\frac{Ï€}{2}$ on the curve $x=a$ $(θ+Sinθ).y=a(1−Cosθ)aî€=1$ ,then

• A. $ST=SN$
• B. $ST=2SN$
• C. $S{T}^2=aS{N}^3$
• D. $S{T}^3=aSN$

Answer 1

Answer: (A)

Given that, $x=a(θ+\sin θ)$ and $y=a(1−\cos θ)$
$⇒\frac{dx}{dθ}=a(1+\cos θ)$ and $\frac{dy}{dθ}=a\sin θ$
$∴\frac{dy}{dx}=\frac{a\sin θ}{a(1+\cos θ)}$
$=\frac{2\sin \frac{θ}{2}\cos \frac{θ}{2}}{2{\cos }^2\frac{θ}{2}}$
$=\tan \frac{θ}{2}$
Now, length of sub tangent $=â\frac{y}{dy/dx}â$
$∴ST=\frac{a(1−\cos θ)}{\tan \frac{θ}{2}}$
$$\begin{gathered} =aâ‹\dots \frac{2{\sin }^2\frac{θ}{2}}{\sin \frac{θ}{2}}â‹\dots \cos \frac{θ}{2} \\ <br/><br/>=a\sin θ \end{gathered}$$
$⇒$ Length of subtangent at $θ=\frac{π}{2}$,
$ST=a\sin \frac{Ï€}{2}=a$
and length of subnormal $=ây\frac{dy}{dx}â$
$⇒SN=a(1−\cos θ)â‹\dots \tan \frac{θ}{2}$
$⇒aâ‹\dots 2{\sin }^2\frac{θ}{2}\tan \frac{θ}{2}$
$⇒$ Length of subnormal at $θ=\frac{π}{2}$,
$SN=aâ‹\dots 2â‹\dots \frac{1}{2}=a$
Hence, $SN=ST$