All points on the curve y2=4ax+asin⁡xa at which the tangents are parallel to the axis of x lie on a

A
circle
B
parabola
C
line
D
none of these

Solution:

We have,

$y^{2} = 4 a (x + a \sin \frac{x}{a})$ ..(i)

$\Rightarrow 2 y \frac{d y}{d x} = 4 a (1 + \cos \frac{x}{a})$

For points at which the tangents are parallel to x-axis, we must have

$\begin{array}{l} \frac{d y}{d x} = 0 \\ \Rightarrow 4 a (1 + \cos \frac{x}{a}) = 0 \Rightarrow \cos \frac{x}{a} = - 1 \Rightarrow \frac{x}{a} = (2 n + 1) π \end{array}$

For these values of $x$ , we obtain $\sin \frac{x}{a} = 0$

Putting $\sin \frac{x}{a} = 0$ , in (i), we get $y^{2} = 4 a x$

Therefore, all these points lie on the parabola $y^{2} = 4 a x .$

All points on the curve $y^{2} = 4 a (x + a \sin \frac{x}{a})$ at which the tangents are parallel to the axis of $x$ lie on a

Solution:

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