Tangent lines to the curve parameterized by $x = a \cos(t)^4$, $y = a \sin(t)^4$

Question

The parametric equations of a curve are $$x = a \cos(t)^4 \qquad y = a \sin(t)^4$$ 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 $$x \sin(t)^2 + y \cos(t)^2 = a \sin(t)^2 \cos(2t) \qquad (3)$$

(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 $-\tan(t)^3 \cot(t)$.

I can't figure out the second part.

Answer 1

for the part a) the answer is slope = $\frac{dy}{dx}$$=$ $\frac{(4a sin^3(t))(cos(t))}{(4acos^3(t))(-sin(t))}$=$\frac{-sin^2(t))}{cos^2(t))}$=$-tan^2(t)$

Answer 2

part a$$x=a\cos^4 t\to \sqrt{\frac{x}{a}}=\cos ^2 t\\ y=a\sin^4 t\to \sqrt{\frac{y}{a}}=\sin ^2 t\\ \sqrt{\frac{y}{a}}+\sqrt{\frac{x}{a}}=\sin ^2 t+\cos ^2t=1\\ \sqrt{\frac{y}{a}}+\sqrt{\frac{x}{a}}=1$$

Answer 3

$$x = a \cos^4t \qquad y = a \sin^4t$$

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

$$OP+OQ = |a|\cos^2(t)+|a|\sin^2(t)=|a|\tag{iii}$$