Implicit Differentiation: Finding points Parallel to y=-x

Question

The Question asks:

Let $x^{2/3}+y^{2/3}=2$ be a curve. Find points on the curve where the tangent line is parallel to $y=-x$.

I have got to $\dfrac{dy}{dx} = -\dfrac{x^{-1/3}}{y^{-1/3}}$ but am confused about how to find the points.

Answer 1

So, at any point $P(h,k)$ on $$x^{2/3}+y^{2/3}=2$$ the gradient will be $$-\dfrac{h^{-1/3}}{k^{-1/3}}$$ which needs to be $-1$

$\implies h=k$ and we have $$2=h^{2/3}+k^{2/3}=2h^{2/3}\implies h^2=1$$

Answer 2

Any point$(P)$ on $x^{2/3}+y^{2/3}=a^{2/3}$

can be written as $P(a\cos^3t,a\sin^3t)$

$$\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dy}{dt}}=-\dfrac{3\cos^2t\sin t}{3\sin^2t\cos t}=-\cot t$$

We need $-\cot t=-1\implies\dfrac{\sin t}1=\dfrac{\cos t}1=\pm\sqrt{\dfrac{\sin^2t+\cos^2t}{1^2+1^2}}=?$