I don't imagine how to make a cartesian equation of the parametric one: $x=2 \sin (t), y=-\frac{4\cos (t)^2 (2 + 1.3\cos(t))}{3 + \sin (t)^2}$

Question

I know it's a downvoting question but I'm not good at math. I know that $x=a \sin(t)$ , $y=\frac{a \cos^2(t)(2+ \cos(t))}{(3- \sin^2(t))}$ is $y^2(a^2-x^2)=(x^2+2ay-a^2)^2$ but what it could be as a cartesian equation with the negative $y$ part I can't even imagine. Sorry. Could you help me?

Answer 1

$$x=2 \sin (t), ~~~~y=-\frac{4\cos (t)^2 (2 + 1.3\cos(t))}{3 + \sin (t)^2}$$ There's a very useful trigonometric identity: $$\sin^2\theta+\cos^2\theta\equiv1$$ where the $\equiv$ symbol (the identity symbol) means the equation is true for all values of $\theta$; that's why it's called an identity. In your case, we know that $$\sin t=\frac{x}{2}$$ Using the above identity, $$\cos ^2t=1-\sin^2 t=1-\frac{x^2}{4}=\frac{4-x^2}{4}$$ Also, $$\cos t=\sqrt{\frac{4-x^2}{4}}=\frac{\sqrt{4-x^2}}{2}$$ (The final equality assumes that $\cos t$ is $+$ve. In fact to be completely correct you should use $\cos t=\pm\frac{\sqrt{4-x^2}}{2}$. Thanks for spotting that @GEdgar.)

Try substituting these values into your equation. If you need more help, don;t hesitate to ask!

Answer 2

$$y=-\frac{4 \cos ^2(t) \left(\frac{13 \cos (t)}{10}+2\right)}{\sin ^2(t)+3}$$ $$\sin t=\frac{x}{2}$$ thus $$y=-\frac{4 \left(1-\frac{x^2}{4}\right) \left(\frac{13 \cos (t)}{10}+2\right)}{\left(\frac{x}{2}\right)^2+3}$$ with some algebra we get $$\cos t=\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}$$ and then $$\cos^2 t=\left(\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}\right)^2$$ as $\cos^2 t= 1-\sin^2 t$ we get $$1-\frac{x^2}{4}=\left(\frac{5 \left(x^2 y-8 x^2+12 y+32\right)}{26 \left(x^2-4\right)}\right)^2$$ and finally $$676 \left(x^2-4\right)^2=25 \left(x^2 y-8 x^2+12 y+32\right)^2+169 x^2 \left(x^2-4\right)^2$$