 Hello friends and how are you all doing today the question says find the slope of the normal to curve x is equal to 1 minus a sine theta y is equal to b cos square theta at theta is equal to power by 2 Here we have x is equal to 1 minus a sine theta and y has b cos square theta now differentiating x and y both with respect to theta we get dx by d theta is equal to minus a cos theta and dy by d theta is equal to minus 2b and derivative of cos theta is minus sine theta so we have 2b cos theta sine theta now by chain rule we have dy by dx equal to dy by d theta into d theta by dx that is the reciprocal of dx by d theta So we have minus 2b cos theta sine theta multiplied by 1 upon minus a cos theta on simplifying we'll have to with 2b upon a sine theta now slow theta equal to pi by 2 is dy by dx at theta equal to pi by 2 will be 2b by a Sine pi by 2 which is 1 so we have the answer as 2b by a That is the slope at theta is equal to pi by 2 But we need to find out the slope of the normal Which is equal to minus 1 upon dy by dx right which is equal to minus 1 upon 2b by a Which is further equal to minus a upon 2b so the answer to the question is minus a upon 2b that is the slope of the norm. This is the required answer to the given question Hope you understood it well and enjoyed it too. Have a nice day