1.x+y+6=0
2.x-y-6=0
3.x-y+6=0
4.x+y-6=0
2.x-y-6=0
3.x-y+6=0
4.x+y-6=0
-
the equation of the normal to the parabola y^2=4ax with slope m is
y=mx-2am-am^3
Here,a=2 (4a=8)
and m=1
therefore the required normal is
y=1.x-4.1-2(1)^3
x-y-6=0
y=mx-2am-am^3
Here,a=2 (4a=8)
and m=1
therefore the required normal is
y=1.x-4.1-2(1)^3
x-y-6=0
-
x-y-6 = 0
-
sex