 Hello and how are you all today? My name is Priyanka and let us discuss the question. It says, find the slope of the normal to the curve x is equal to a cos cube theta, y is equal to a sin cube theta, at theta is equal to pi by 4. Now before proceeding on with the solution, let us be well versed that a normal is perpendicular to the tangent, right? So the slope of the normal to the curve y is equal to fx at x0, y0 is minus 1 upon f dash x0 if f dash x0 is not equal to 0, right? The knowledge of this is the key idea for this question. And now on using this key idea, we will be proceeding with the solution. Now here we have x equal to a cos cube theta and we have y is equal to a sin cube theta. We will be differentiating x and y with respect to theta. We have dx by d theta as 3a cos square theta into minus sin theta and we have dy by d theta as 3a sin square theta into cos theta. Now further by chain rule, we can easily find out dy by dx that is on multiplying dy by d theta by d theta by dx that is the reciprocal of dx by d theta. So we have 3a sin square theta cos theta multiplied by 1 upon 3a cos square theta into minus sin theta. Now on simplifying, we have minus sin theta upon cos theta which can be written as minus tan theta. Now we need to find out the slope at which theta is equal to pi by 4, right? So we need to find out the value of dy by dx where theta is pi by 4 which is equal to minus tan pi by 4 which is equal to minus 1. We know that since slope of the normal is minus 1 upon dy by dx so that means it will be equal to minus 1 upon minus tan pi by 4 whose value we have found out above as minus 1. So on further simplifying, we have our answer as 1. So 1 is the answer for this question. Hope you understood it well and enjoyed it too. Have a nice day.