Vtyjkyuk

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:

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

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

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!

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

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

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

Reducing we get:

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

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

$a =frac14$

$b=-frac12$

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

My work:
$$y^2+ax^2+bx=0$$
Substituting the values of $x$ and $y$ we get,
beginarrayll
&sin^2(t).cos^2(t)+4asin^4(t)+2bsin^2(t)&=0\
&implies sin^2(t)(1-sin^2(t))+4asin^4(t)+2bsin^2(t)&=0\
&implies sin^2(t)-sin^4(t)+4asin^4(t)+2bsin^2(t)&=0\
&implies 2bsin^2(t)+sin^2(t)&=sin^4(t)-4asin^4(t)\
&implies sin^2(t)(2b+1)&=sin^4(t)(1-4a)\
&implies dfrac12.2sin^2(t)(2b+1)&=dfracbig(2sin^2(t)big)^24(1-4a)\
&implies dfracx2(2b+1)&=dfracx^24(1-4a)\
&implies x(4b+2)-x^2(1-4a)&=0\
&implies x(4b+2-x+4ax)&=0\
endarray
$$textThis gives,$$

$$x=0 or, x= dfrac4b+21-4a$$
Now, let's think what does this mean.if $x= dfrac4b+21-4a$ then, what are the conditions?
$$here,(1-4a)ne0$$
$$implies ane dfrac14$$
Hence, $b in mathbbR$ and $ain mathbbR - dfrac14$

Again if we take $x=0$ ,then $a=dfrac14$ and $b=-dfrac12$