The coordinate of the point p on the curve y2=2x3,  the tangent at which is perpendicular to the line 4x−3y+2=0,  are given by

Differentiating y2=2x3,  we get 2ydydx=6x2⇒dydx=3x2y The slope of the line 4x−3y+2=0  is 4/3.  Therefore , the coordinate of p(x,y)  must satisfy 3x2y.43=−1⇒4x2=−y Also , y2=2x3. Solving these , we get x=1/8  and  y=−1/16  (clearly  x≠0).