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I came across the following question:

The coordinates of a particle moving in a plane are given by $x(t)=a\cos(pt)$ and $y(t)=b\sin(pt)$, where $a>b$ and $a$ and $b$ are positive constants of appropriate dimensions.

This is not the question being asked but what condition should the acceleration(given by differentiating $x(t)$ and $y(t)$ twice with respect to time $t$) satisfy for us to state that it's always directed towards the focus of the ellipse $x^2/a^2 + y^2/b^2$ that the particle traces?

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  • $\begingroup$ Have you tried differentiating to find the components of acceleration, $a_x=\ddot x$ and $a_y=\ddot y$? $\endgroup$
    – sammy gerbil
    Commented Aug 19, 2016 at 3:31
  • $\begingroup$ Yes, I have; ax= -p²acos(pt) and ay= p²bsinpt. How to proceed with this information, to show that acceleration always points toward the focus of the ellipse traced by the particle? Also, why did u downvote this question? $\endgroup$
    – user106570
    Commented Aug 19, 2016 at 3:41
  • $\begingroup$ I down-voted because you have not shown effort to solve your own problem. See the site policy for such exercises ... Elliptical motion results from a central force (ie directed towards a point) of the form $F=ma=k/r^2$. The components ($a_x, a_y$) give you the direction of the acceleration $a$ at the point ($x, y$). Can you use the geometry of the ellipse to show that $a$ points towards one focus? And is proportional to $1/r^2$ where $r$ is distance from that focus? $\endgroup$
    – sammy gerbil
    Commented Aug 19, 2016 at 4:08
  • $\begingroup$ Technically, mine isn't a homework question since it was related to a concept and not exactly specific to this problem alone. Furthermore, I'm sorry that I didn't include the values of acceleration that I had found by differentiating in the question itself; I will keep this in mind when posting questions in the future. I was(and still am) confused about how to attempt to prove that a given acceleration vector always points toward a certain point. Can u please explain some more? $\endgroup$
    – user106570
    Commented Aug 19, 2016 at 4:31
  • $\begingroup$ Oh crap, the acceleration comes out to be ax=-p²(acospt) and at=-p²(bsinpt). Both are negative. My bad, sorry. $\endgroup$
    – user106570
    Commented Aug 19, 2016 at 4:38

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Note sure if I understand your question, but I'll try to give an answer. The acceleration is proportional to the force thanks to $\mathbf{F}=m\mathbf{a}$. So if the acceleration point to the center (focus of the ellipse}, then it is because the force points to the center. The force can (under certain conditions) be derived from a potential $\mathbf{F}=-\nabla \phi$. So if the force points to the center, then the associated potential is symmetric around the center and it increases as one moves away from the center.

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  • $\begingroup$ Can u please explain the second equation in ur answer? I'm afraid that I'm not aware of it(I only just graduated high school). $\endgroup$
    – user106570
    Commented Aug 19, 2016 at 4:40
  • $\begingroup$ Actually, after differentiating, the acceleration vector can be written in terms of the position vector as a=-p²r, where r is the position vector r=xi+yj. This means that for every point on the ellipse, the acceleration points in fact, toward the origin; am I correct in assuming that? $\endgroup$
    – user106570
    Commented Aug 19, 2016 at 4:42
  • $\begingroup$ Potentials are things one learns about, when you learn about Hamiltonians and stuff. That stuff is quite a mouth full. If you are interested in that you can try to find some introductory text on classical mechanics. I love your enthusiasm. $\endgroup$
    – flippiefanus
    Commented Aug 19, 2016 at 4:47
  • $\begingroup$ @KaumudiHarikumar : Yes, you are correct. If $t$ is time then the acceleration vector is directed towards the centre of the ellipse, not one focus. The law of force in this case is $F=ma= -mp^2r$. $\endgroup$
    – sammy gerbil
    Commented Aug 19, 2016 at 14:40