49.xy12x2y50}ddx}(xy)1}ddx}(2x)2}ddx}(y)50x}ddyx}1(y)(1)122}ddyx}50(x21)}ddyx}5 222y}ddyx}5}2x2221y}5}2112yx}Since the slope of the line 2x1y5 0 is 22, we wish tofind points where the normal has slope22, that is, wherethe tangent has slope }12}. Thus, we have}2112yx}5}12}2(21y)512x412y512xx5 22y23Substituting 22y23 in the original equation, we have:xy12x2y50(22y23)y12(22y23)2y5022y228y265022(y11)(y13)50y5 21 or y5 23At y5 21,x5 22y2352235 21.At y5 23:x5 22y23562353.The desired points are (21,21) and (3,23).Finally, we find the desired normals to the curve, which arethe lines of slope22 passing through each of these points.At (21,21), the normal line is y5 22(x11)21 or y5 22x23. At (3,23), the normal line is y5 22(x23)23 or y5 22x13.50.x5y2}ddx}(x)5}ddx}(y2)152y}ddyx}}ddyx}5}21y}The normal line at (x,y) has slope 22y. Thus, the normal
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