Tangent lines to the curve parameterized by $x = a \cos^4t$, $y = a \sin^4t$ The parametric equations of a curve are
$$x = a \cos^4t \qquad y = a \sin^4t$$
where $a$ is a positive constant.
(i) Express $\dfrac{dy}{dx}$ in terms of $t$. (3)
(ii) Show that the equation of the tangent to the curve at the point with parameter $t$ is
$$x \sin^2t + y \cos^2t = a \sin^2t \cos2t \qquad (3)$$
(iii) Hence show that if the tangent meets the $x$-axis at $P$ and the $y$-axis at $Q$, then
$$|OP| + |OQ| = a$$
where $O$ is the origin. (2)
My answer to the first part is $-\tan^3t \cot t$.
I can't figure out the second part.
 A: for the part a) the answer is slope = $\frac{dy}{dx}$$=$ $\frac{(4a sin^3(t))(cos(t))}{(4acos^3(t))(-sin(t))}$=$\frac{-sin^2(t))}{cos^2(t))}$=$-tan^2(t)$
A: part a$$x=a\cos^4 t\to \sqrt{\frac{x}{a}}=\cos ^2 t\\
y=a\sin^4 t\to \sqrt{\frac{y}{a}}=\sin ^2 t\\
\sqrt{\frac{y}{a}}+\sqrt{\frac{x}{a}}=\sin ^2 t+\cos ^2t=1\\
\sqrt{\frac{y}{a}}+\sqrt{\frac{x}{a}}=1$$
A: $$x = a \cos^4t \qquad y = a \sin^4t$$
This is what I got.
$$\dfrac{dy}{dx}
   =\dfrac{4a \sin^3(t) \cos(t)}{-4a \cos^3(t) \sin(t)}
   =-\dfrac{\sin^2(t)}{\cos^2(t)}\tag{i}$$
\begin{align}
   y-a\sin^4(t) &= -\dfrac{\sin^2(t)}{\cos^2(t)}(x-a\cos^4(t)) \\
   y\cos^2(t) - a\sin^4(t)\cos^2(t) &= -x\sin^2(t) + a\cos^4(t)\sin^2(t)\\
   x\sin^2(t) + y\cos^2(t) &=a\sin^4(t)\cos^2(t)+a\cos^4(t)\sin^2(t)\\
   x\sin^2(t) + y\cos^2(t) &=a\sin^2(t)\cos^2(t)[\sin^2(t)+\cos^2(t)]\\
   x\sin^2(t) + y\cos^2(t) &=a\sin^2(t)\cos^2(t)\\
   \dfrac{x}{a\cos^2(t)}+\dfrac{y}{a\sin^2(t)} &= 1\tag{ii}
\end{align}
Hence 
the $x$-intercept is $P=(a\cos^2(t),0)$ and 
the $y$-intercept is $Q=(0,a\sin^2(t))$. So
$$OP+OQ = |a|\cos^2(t)+|a|\sin^2(t)=|a|\tag{iii}$$
