The slope of the normal to the curve x3+y3=8xy  at the point, other then origin, where it meets the curve y2=4x  is__________.

Then curves are x3+y3=8xy …..(1)  and y2=4x …..(2) Solving (1)  and (2),  we get  x3+y3=8xy                       ⇒x3=4xy                          ⇒x3=4x2x                       ⇒x3/2(x3/2−8)=0                        ⇒x=0 or x=22=4 Now when x=0,   we get y=0 And when x=4,  we get y2=16 0r y±4. But x=4  and y=−4  do not satisfy  (1). Thus (0,0)  and (4,4)  are the points of intersection of(1)  and  (2) Differentiating (1) , we get dydx=8y−3x23y2−8x. at(4,4),dydx=−1. Hence, the equation of the normal to the curve (1)  at (4,4)  is  (y−4)=1(x−4)or  y−x=0.