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# if x+y=k is normal to y^2=12x,then k is

10 years ago

We know the slope of the normal which is (-1). This implies that the slope of the tangent is 1. Now,

y= mx+c is a tangent to y^2=4ax, iff. c=a/m.

Putting a=3, m=1, we get c=3.

Tangent is y=x+3.

Solving the equations of parabola and tangent, we get pts (3,6) and(3,-6)

This means that k=-3 or k=9

10 years ago

x + y = k

y = -x + k

slope of normal = -1

so slope of tangent = 1

parabola is y2 = 12x          ..............1

2yy1 = 12

y1 = 1 so  , y = 6

from eq 1 , if y = 6 then x = 3

x+y = k touches parabola at (3,6) so it will satisfy this eq

at (3,6) , k = 9

therefore value of k is 9

3 years ago
Here a=3
We know that equation of normal is
Y=mx-am^3-2am
Comparing the equation x+y=k from equation of normal
So k= am+2am
K=3+6 (putting value of a and m=1)
K=9
K=3+6 (putting value of a and slope is 1)
So k=9
2 years ago
Normal with slope m to the parabola y*y =12x Is
Y =mx-2am-am3(cube)
If c=-2am-am3(cube)
After comparison a=3,m=1
K=-2*3*1-3*1*1*1=-6-3 =-9
one year ago
Dear Student,

x + y = k
y = -x + k
slope of normal = -1
so slope of tangent = 1
parabola is y2 = 12x ..............1
2yy1 = 12
y1 = 1 so , y = 6
from eq 1 , if y = 6 then x = 3
x+y = k touches parabola at (3,6) so it will satisfy this eq
at (3,6) , k = 9
therefore value of k is 9

Thanks and Regards