Grade 11Analytical Geometry

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

ashok kumar

15 Years agoGrade 11

5 Answers

Shreyansh Gandhi

Approved Tutor Answer15 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

vikas askiitian expert

Approved Tutor Answer15 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

Shashank raj

7 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

anildualecturer

7 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

Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

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