The coordinates 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=3x2yThe slope of the line 4x – 3 y+2 =0 is 4/3.  Therefore , the coordinates 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)