Find the interval for z so that (3-Z)X+ZY+(z2-1)=O is normal to the curve XY=4

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prince sunjot dutt · Answer

(3-z)x+zy+(2z-1)=0 zy=(z-3)x+(1-2z) Y={(z-3)/z}x+(1-2z)/z From, Y=mX+c m=(z-3)/z. [SLOPE OF NORMAL]............(1) Given curve; xy=4 y=4/x dy/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^2/4 1-3/z=x^2/4 3/z=(4-x^2)/4 z=12/4-x^2 For z to be defined X^2=|= 4 x=|= +-2 Interval z€R-{+-2}

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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

Find the interval for z so that (3-Z)X+ZY+(z2-1)=O is normal to the curve XY=4

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