If the line x + By + C = 0 is the normal to the curve given by x = a ^3 t , y = b ^3 t , (where a, b 0 ) at a point t = 2 , then B - C =

Step-by-step solution

Given the parametric equations of the curve: x = a ^3 t y = b ^3 t At t = 2 , the coordinates of the point on the curve are: x = a ^3 ( 2 ) = a(1)^3 = a y = b ^3 ( 2 ) = b(0)^3 = 0 So, the point is (a, 0) . Differentiating x and y with respect to t : dx dt = 3a ^2 t t dy dt = -3b ^2 t t The slope of the tangent to the curve is: dy dx = dy dt dx dt = -3b ^2 t t 3a ^2 t t = - b a t At t = 2 , the slope of the tangent is dy dx = - b a ( 2 ) = 0 . Since the tangent is horizontal, the normal to the curve is vertical (pa