$$x = \sin^2(t), y = \cos(t)$$ I know that to eliminate parameter involving $\sin$ and $\cos$, we should reduce it to $x^2 + y^2 = r^2$.
So, $x^2 = \sin^4(t)$, $y^2 = \cos^2(t)$
But I can't make $\sin^4(t)$ and $\cos^2(t)$ to $1$.
Any tips?
$$x = \sin^2(t), y = \cos(t)$$ I know that to eliminate parameter involving $\sin$ and $\cos$, we should reduce it to $x^2 + y^2 = r^2$.
So, $x^2 = \sin^4(t)$, $y^2 = \cos^2(t)$
But I can't make $\sin^4(t)$ and $\cos^2(t)$ to $1$.
Any tips?
The Cartesian equation you're looking for here is $x+y^2=1$.
I think you should use $\sin^2(t) + \cos^2(t) = 1$. Then $$ 1 = \sin^2(t) + \cos^2(t) = x + y^2 \, . $$
Notice that if you just consider $x+y^2$ then you have $\sin^2(t)+\cos^2(t)=1$.
This means you have the equation $x=1-y^2$. The shape, then, is a parabola with vertex at $(1,0)$ and $y$-intercepts at $(0, 1)$ and $(0, -1)$. In other words, it's much like your typical parabola $y=1-x^2$, only its graph has been rotated and shifted a bit!