prince sunjot dutt
Last Activity: 6 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}