A curve is given parametrically by the equations
$x = a\cos^3 t\ ,\ y = a\sin^3t \ $ $ \ 0\le t \le 2\pi$
Where a is a positive constant. Show that the equation of the tangent at the point with parameter p where $\sin2p \neq 0$ is given by
$x\sin p +y\cos p = a \ \sin p \ \cos p$
A second tangent to the curve, perpendicular to the first tangent, intersects the first tangent at the point with polar coordinates ($r,\theta$). Show that $2r^2 = a^2\sin^2(2p)$ and deduce that the polar equation of the locus of the points from which a pair of perpendicular tangents can be drawn to the curve is $2r^2 = a^2\cos^2(2\theta)$
I had attempt first first part of the problem which it is proven the tangent equation is as given $x\sin p +y\cos p = a \ \sin p \ \cos p$.
However, I was unable to prove $2r^2 = a^2\sin^2(2p)$ by getting the tangent perpendicular to the first tangent.
My attempt : $\frac{dy}{dx} =-\tan\ p $, $\frac{-dx}{dy} =\cot\ p $
A : Point at which first tangent touches the curve.
B : Point at which second tangent touches the curve.
Since we know A ($a\cos^3 p\ ,\ a\sin^3p$), then B must either have sin or cos being negative and B($-a\cos^3 p\ ,\ a\sin^3p$). I am not very sure with this deduction which I expands and unable to prove the statement above.
As you said the slope at A is $-\tan p$ so the slope at B should be $\cot p$ then what is the value of $t$ such that $-\tan t =\cot p$ is it $p+\frac{\pi}{2}$? Find the tangent line then find the point of intersection the convert to polar