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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

Grade:12th pass

1 Answers

prince sunjot dutt
58 Points
3 years ago
(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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