Prove that all the points on the curve $$y^2 = 2a(x+a\sin \frac{x}{a})$$ at which tangent is parallel to the axis of $x$, lie on a parabola.
Here slope of tangent at $(h,k)$ must be $0$. After calculating $\frac{dy}{dx}$ at $(h,k)$ and equating it to $0$, I obtained $a=\frac{h}{\pi}$. On putting value of $a$ in $y^2 = 2a(x+a\sin \frac{x}{a})$, I am getting equation of pair of straight lines and not an equation of parabola. What am I doing wrong?
Implicit differentiation
$$2yy'=2a\left(1+\cos\frac xa\right)=0\iff y'=\frac{a\left(1+\cos\frac xa\right)}{y}\iff$$
$$\cos\frac xa=-1\iff x=a(2k-1)\pi\,,\;\;k\in\Bbb Z$$
and then substituting in original equation
$$y^2=2a\left(a(2k-1)\pi+a\,\overbrace{\sin\frac{a(2k-1)\pi}a}^{=0}\right)=2a^2(2k-1)\pi=2ax$$
a parabola