Q:

Tangent and Normal

Parametric equation of given curve is $x=a(2\cos t+\cos 2t)$,
$y=a(2\sin t-\sin 2t)$ . I need to find equation of normal at any point $t$.
I have found out $-\frac{dx}{dy}$ for the curve which turns out to be $-\cot(t/2)$ I am unable to proceed after this since only one point is mentioned. I think the normal form of a line concept is to be used. Maybe I am wrong. Please solve the question.

A:

Hint:
I suppose we have $a\ne 0$ (otherwise, this is not really a curve). The derived vector is $\;\bigl(-2a(\sin t+\sin 2t), 2a(\cos t-\cos 2t)\bigr)$, which is collinear to $\;(-\sin t-\sin 2t,  \cos t-\cos 2t)$.
Thus, as long as the derived vector is not $0$, a normal vector of the curve at the point with parameter $t$ is
$$\vec n=(\cos t-\cos 2t, \,\sin t+\sin 2t). $$
The vector equation of this normal is
$$\vec n\cdot (x,y)=\vec n\cdot\bigl(x(t),y(t)\bigr).$$
Can you take it from there?

