Solve for two variables of a parametric function

Question

There are constants $a$ and $b$ so that for all points (x, y) on curve K it holds: $y^2+ax^2+bx=0$

Curve K is presented by the parametric equation:

$x(t)=2\sin^2(t)$

$y(t)=\sin(t)\cos(t)$

find $a$ and $b$

Looking for a way to solve this, I tried substituting $y(t)$ and $x(t)$ into the parabolic equation but nothing makes sense then.

Explanation is really appreciated! Thanks!

Answer 1

$$\sin^2t\cos^2t+4a\sin^4t+2b\sin^2t=u(1-u)+4au^2+2bu=0$$ with $a=\frac14,b=-\frac12$.

Answer 2

$\cos 2t=1-2\sin^2 t=1-x$

$2y = 2\sin t \cos t=\sin 2t$

$\sin^2 2t + \cos^2 2t =(1-x)^2 +(2y)^2=1$

Reducing we get:

$4y^2+x^2-2x=0$

$y^2 +(1/4)x^2+(-2/4)x=0$

$a =\frac{1}{4}$

$b=-\frac{1}{2}$

Answer 3

My work: $$y^2+ax^2+bx=0$$ Substituting the values of $x$ and $y$ we get, \begin{array}{ll} &\sin^2(t).\cos^2(t)+4a\sin^4(t)+2b\sin^2(t)&=0\\ &\implies \sin^2(t)(1-\sin^2(t))+4a\sin^4(t)+2b\sin^2(t)&=0\\ &\implies \sin^2(t)-\sin^4(t)+4a\sin^4(t)+2b\sin^2(t)&=0\\ &\implies 2b\sin^2(t)+\sin^2(t)&=\sin^4(t)-4a\sin^4(t)\\ &\implies \sin^2(t)(2b+1)&=\sin^4(t)(1-4a)\\ &\implies \dfrac{1}{2}.2\sin^2(t)(2b+1)&=\dfrac{\big(2\sin^2(t)\big)^2}{4}(1-4a)\\ &\implies \dfrac{x}{2}(2b+1)&=\dfrac{x^2}{4}(1-4a)\\ &\implies x(4b+2)-x^2(1-4a)&=0\\ &\implies x(4b+2-x+4ax)&=0\\ \end{array} $$\text{This gives,}$$

$$x=0 or, x= \dfrac{4b+2}{1-4a}$$ Now, let's think what does this mean.if $x= \dfrac{4b+2}{1-4a}$ then, what are the conditions? $$here,(1-4a)\ne0$$ $$\implies a\ne \dfrac{1}{4}$$ Hence, $b \in \mathbb{R}$ and $a\in \mathbb{R} - {\dfrac{1}{4}}$

Again if we take $x=0$ ,then $a=\dfrac{1}{4}$ and $b=-\dfrac{1}{2}$