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**arze** A curve is given parametrically by

$\displaystyle x=a(5\cos\theta+\cos 5\theta)$

$\displaystyle y=a(5\sin\theta-\sin 5\theta)$

Find the equation of the normal to the curve at the point with parameter $\displaystyle \theta$.

Find those points on the curve at which the normal is also normal to the curve at another point.

my attempt:

Gradient

$\displaystyle \frac{dx}{d\theta}=-5a(\sin\theta+\sin 5\theta)$

$\displaystyle \frac{dy}{d\theta}=5a(\cos\theta-\cos 5\theta)$

$\displaystyle \frac{dy}{dx}=-\frac{\cos\theta-\cos 5\theta}{\sin\theta+\sin 5\theta}$

normal

$\displaystyle \frac{\sin\theta+\sin 5\theta}{cos\theta-\cos 5\theta}=-\frac{2\sin 3\theta\cos 2\theta}{2\sin 3\theta\sin 2\theta}=-\frac{\cos 2\theta}{\sin 2\theta}$

equation

$\displaystyle y-a(5\sin\theta-\sin 5\theta)=-\frac{\cos 2\theta}{\sin 2\theta}(x-a(5\cos\theta+\cos 5\theta))$

$\displaystyle y\sin 2\theta-a\sin 2\theta(5\sin\theta-\sin 5\theta)=-x\cos 2\theta+a\cos 2\theta(5\cos\theta+\cos 5\theta)$

$\displaystyle y\sin 2\theta+x\cos 2\theta=a[5\cos\theta\cos 2\theta+\cos 2\theta\cos 5\theta+5\sin 2\theta\sin\theta-\sin 2\theta\sin 5\theta]$

$\displaystyle y\sin 2\theta+x\cos 2\theta=$$\displaystyle a[\frac{5}{2}(\cos 3\theta+\cos\theta)+\frac{1}{2}(\cos 7\theta+\cos 3\theta)-\frac{5}{2}(\cos 3\theta-\cos\theta)+\frac{1}{2}(\cos 7\theta-\cos 3\theta)]$

$\displaystyle y\sin 2\theta+x\cos 2\theta=a(5\cos\theta+\cos 7\theta)$

The answer is: $\displaystyle y\sin 2\theta+x\cos 2\theta=a(5\cos\theta-\cos 3\theta)$

So I got the RHS correct, but where on the LHS is wrong?

And for finding the points where the normal there is normal at another point, I don't know where to start.

Thanks for any help