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Differential Calculus
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20 Sep 2007 21:52:37 IST
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i' tell you the procedure....
first find out dy/dx for the curve.....[ dy/dx = (dy/dt) / (dx/dt) ]
take the slope at (x1,y1) say....
Now write the equation for normal at (x1,y1)....
as y - y1 = {-1 / (dy/dx) } (x - x1)
Now use the formula for perpendicular distance on a line from a point (here (0,0) ) to get the reqd. dist = a
first find out dy/dx for the curve.....[ dy/dx = (dy/dt) / (dx/dt) ]
take the slope at (x1,y1) say....
Now write the equation for normal at (x1,y1)....
as y - y1 = {-1 / (dy/dx) } (x - x1)
Now use the formula for perpendicular distance on a line from a point (here (0,0) ) to get the reqd. dist = a



R+)









eqn of normal is y-asint +acost=-cot t(x-acos t - atsin t)
we get y+cot t-asin t+acos tcot t=0
distance from origin=a