Answer: (i) \(r=-2a\sin2\theta\). The curve is a rose curve with loops as shown by this polar form.
(ii) \((x^2+y^2)^{3/2}=-4axy\).
(iii) The area of one loop is \(\frac12\pi a^2\).
(iv) At the required points, \(2\tan\theta=-\tan2\theta\).
Using the angle-addition formula,
\(\cos\left(2\theta+\frac{\pi}{2}\right)=\cos2\theta\cos\frac{\pi}{2}-\sin2\theta\sin\frac{\pi}{2}\).
Therefore
\(r=2a\cos\left(2\theta+\frac{\pi}{2}\right)=2a(0-\sin2\theta)=-2a\sin2\theta\).
The sketch is the corresponding polar rose. One loop lies in the second quadrant and another in the fourth quadrant, with the appropriate symmetry through the pole.
For the Cartesian form, use \(x=r\cos\theta\) and \(y=r\sin\theta\). Since
\(r=-2a\sin2\theta=-4a\sin\theta\cos\theta\),
we get
\(r=-4a\frac{y}{r}\frac{x}{r}\).
Thus
\(r^3=-4axy\).
Since \(r^2=x^2+y^2\), this becomes
\((x^2+y^2)^{3/2}=-4axy\).
For one loop, take \(\frac{\pi}{2}\le\theta\le\pi\). The polar area formula gives
\(A=\frac12\int_{\pi/2}^{\pi}r^2\,d\theta\).
So
\(A=\frac12\int_{\pi/2}^{\pi}4a^2\sin^22\theta\,d\theta=2a^2\int_{\pi/2}^{\pi}\sin^22\theta\,d\theta\).
Using \(\sin^22\theta=\frac12(1-\cos4\theta)\),
\(A=a^2\int_{\pi/2}^{\pi}(1-\cos4\theta)\,d\theta\).
Hence
\(A=a^2\left[\theta-\frac14\sin4\theta\right]_{\pi/2}^{\pi}=\frac12\pi a^2\).
Finally, \(y=r\sin\theta=-2a\sin2\theta\sin\theta\).
A tangent parallel to the initial line has \(\frac{dy}{d\theta}=0\), at points where the curve is not at the pole.
Differentiate:
\(\frac{dy}{d\theta}=-2a(2\cos2\theta\sin\theta+\sin2\theta\cos\theta)\).
Setting this equal to zero gives
\(2\cos2\theta\sin\theta=-\sin2\theta\cos\theta\).
Dividing by \(\cos\theta\cos2\theta\), where this is valid for the non-pole tangent points, gives
\(2\tan\theta=-\tan2\theta\).
