Let the normal at a point P on the curve y2-3x2+y+10=0 intersect the y -axis at 0,32. If m is the slope of the tangent at P to the curve, then |m| is equal to

Let P=x1,y1.   Then equation of tangent at P is2yy'-6x+y'=0⇒y'=6x11+2y1∴slope of normal at P=32-y1-x1=-1+2y16x1⇒9-6y1=1+2y1 ⇒y1=1∴x1=±2∴ Slope of tangent =m=±123                                    =±4∴m=4