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```				   26. 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 (< a) and p are positive constants of appropriate dimensions. Then : (A) the path of the particle is an ellipse (B) the velocity and acceleration of the particle are normal to each other at t = p/2p (C) the acceleration of the particle is always directed towards a focus (D) the distance travelled by the particle in time interval t = 0 to t = p/2p is a.
my doubt is that option c should be wrong... in the answer key its given to be true which says that the focus is the centre which in no way is visible to me plz confirm..
```

5 years ago

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```										x = acospt                         y = bsinpt
R(position vector) = acospt(i) + bsinpt(j)        ..................1

dx/dt = vx = -apsinpt           dy/dt = vy =  bpcospt
v = -apsinpt(i) + bpcospt (j)

dv/dt = ax = -ap2cospt         dv/dt = ay = -bp2sinpt
a = -ap2cospt(i) - bp2sinpt(j)
a = -p2( acospt (i) + bsinpt (j) )      .....................2
from eq 1
a = -p2(R)
R is always directrd towards center (0,0) so accleration
of particle will always be towards center of ellipse ....
```
5 years ago
```										This Question has more than 1 answer
x =  a cos pt => cos pt = x/a
y = b sin pt => sin pt = y/b
squaring and adding them we get
x2/a2 + y2/b2 = 1
Therefore path is an ellipse --------------- (option A is correct)
dx/dt = vx = -a p sin pt
d2x/ dt2 = ax = - ap2 cos pt
dy/dt = vy = bp cos pt
d2y/dt2 = - bp2 sin pt
At time t = pi/2p or pt = pi/2
ax and vy become 0
only vx and vy are left
or we can say velocity is along the negative x axis and accelaration along the negative y axis
at t = pi/2p , velocity and accelaration are normal to each other , so option B is also correct.
At t=t , position of the particle
r-> = x i^ + y j^ = a cos pt i^ + b sin pt  j^
and accelaration of the particle is
a-> (t) = ax i^ + ay j^
= - p2 [ a cos pt i^ + b sin pt j^ ]
= - p2 [ x i^ + y j^ ]
= - p2 r-> (t)
so the accelaration of the particle is always directed towards origin
hence option (C) is also correct

```
5 years ago
```										my dear sudhesh the focus of the ellipse is not the center and that is what i wish to highlgiht..
```
5 years ago
```										Dear Mayank ,
if you read carefully i said it is towards the origin not the focus , so , i m really sorry , option (C) is wrong , i thought the question was about particle always directed towards the origin , i didn't see the 'focus' part.

so option (A) is correct and option (B) is correct.

```
5 years ago
```										yaar sudhesh option c says accn. is directed towards a focus wheras uve proved it to be directed towards te fosuc and seeing the trajectory of particle the focus in no way is the center of the ellipse...
```
5 years ago
```										x =  a cos pt => cos pt = x/ay = b sin pt => sin pt = y/bsquaring and adding them we getx2/a2 + y2/b2 = 1Therefore path is an ellipse --------------- (option A is correct)dx/dt = vx = -a p sin ptd2x/ dt2 = ax = - ap2 cos ptdy/dt = vy = bp cos ptd2y/dt2 = - bp2 sin ptAt time t = pi/2p or pt = pi/2ax and vy become 0only vx and vy are leftor we can say velocity is along the negative x axis and accelaration along the negative y axisat t = pi/2p , velocity and accelaration are normal to each other , so option B is also correct.At t=t , position of the particler-> = x i^ + y j^ = a cos pt i^ + b sin pt  j^and accelaration of the particle isa-> (t) = ax i^ + ay j^= - p2 [ a cos pt i^ + b sin pt j^ ]= - p2 [ x i^ + y j^ ]= - p2 r-> (t)so the accelaration of the particle is always directed towards originhence option (C) is also correct
```
2 years ago

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