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Find the interval for z so that (3-Z)X+ZY+(z2-1)=O is normal to the curve XY=4
(3-z)x+zy+(2z-1)=0zy=(z-3)x+(1-2z)Y={(z-3)/z}x+(1-2z)/zFrom, Y=mX+cm=(z-3)/z. [SLOPE OF NORMAL]............(1)Given curve;xy=4y=4/xdy/dx= -4/x^2. [SLOPE OF TANGENT TO CURVE]SLOPE OF NORMAL= -1/(dy/dx)=x^2/4. …...............(2)(1)=(2)z-3)/z=x^2/41-3/z=x^2/43/z=(4-x^2)/4z=12/4-x^2For z to be defined X^2=|= 4x=|= +-2Interval z€R-{+-2}
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