If the tangent to the curve x = a t2, y = 2 at is perpendicular to x-axis, then its point of contact is _____________ .

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(0, 0) Let the required point be (x1, y1). 	ext { Since, the point lies on the curve } . 	ext { Hence, } x_1 = a t^2 	ext { and } y_1 = 2	ext { at } 	ext { Now }, x = a t^2 	ext { and } y = 2	ext { at } \Rightarrow dx/dt = 2	ext { at and } dy/dt = 2a \Rightarrow dy/dx = dy/dt/dx/dt = 2a/2at = 1/t = 2a/y 	ext { Slope of the tangent } = \left( dy/dx ight)_\left( x_1 , y_1 ight) =2a/y_1 	ext { It is given that the tangent is perpendicular to the y-axis. } 	ext { It means that it is parallel to thex-axis }. 	herefore 	ext { Slope of the tangent = Slope of the x-axis } 2a/y_1 = 0 \Rightarrow a = 0 	ext { Now }, x_1 = a t^2 = 0 	ext { and } y_1 = 2	ext { at }= 0 	herefore \left( x_1 , y_1 ight) = \left( 0, 0 ight)

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