 Hello and welcome to the session. Let us understand the following question which says a particle moves along the curve 6y is equal to xq plus 2. Find the points on the curve at which the y coordinate is changing 8 times as fast as the x coordinate. Now, let us proceed on to the solution. Given to us the equation of curve as 6y is equal to xq plus 2 and also it is given to us that y coordinate is changing 8 times fast as x coordinate which implies dy by dt is equal to a multiplied by dx by dt. Now, consider the given equation of curve that is 6y is equal to xq plus 2 on differentiating it with respect to t on both the sides we get 6dy by dt is equal to 3x square dx by dt. Now, substituting the value of dy by dt from 1 into we get 6 multiplied by 8 dx by dt is equal to 3x square dx by dt. Now, dx by dt and dx by dt gets cancelled. So, it implies 8 sigma 48, 48 is equal to 3x square. This implies x square is equal to 48 divided by 3. 48 gets cancelled by 3 and we get here 16 which implies x is equal to plus minus 4. Now, when x is equal to plus 4, substituting in the given curve we get 6y is equal to 446, 16464 plus 2 that is 6y is equal to 66 and it implies y is equal to 11. Hence, the first required point is 4 comma 11. Now, when x is equal to minus 4, 6y is equal to minus 64 plus 2. It implies 6y is equal to minus 62. It implies y is equal to minus 62 by 6 which is equal to minus 33, 31 by 3. Hence, the second required point is minus 4 comma minus 31 by 3. Hence, the required points are 4 comma 11 and minus 4 comma minus 31 by 3. This is the required answer. I hope you understood this question. Bye and have a nice day.