Question

(i) Show that for all values of m, the line y = mx − 3m2 touches the parabola
x2 = 12y.
(ii) Find the values of m for which this line passes through the point (5, 2).
(iii) Hence determine the equations of the two tangents to the parabola
x2 = 12y from the point (5, 2). Anyone know how to do this ?

Answer 1

any point on the parabola (x^2=4*3*y) is of the form (6t,3t^2)
differntiating x^2=12y we have 2x=12(dy/dx)
or dy/dx =x/6
at (6t,3t^2) dy/dx =6t/6=t (slope of all the tangents passing through (6t,3t^2) on the parabola)
hence eq of tangent through (6t,3t^2) is of the form
then y-3t^2=t(x-6t)
or the eq of tangent is y=tx +3t^2-6t^2
or y=tx -3t^2
that is to say for all real values of 't' y=tx-3t^2 is always tangent to the x^2=12y
hence if t=m
then y=mx-3m^2 is always tangent to the given parabola (procved)

Answer 2

if y=mx-3m^2 passes through (5,2)
then
2=5m-3m^2
or 3m^2-5m+2=0
or 3m^2-3m-2m+2=0
or 3m(m-1)-2(m-1)=0
or (m-1)(3m-2)=0
thus the vlues of m are 1 and 2/3
and for m=1
tangent is y=x-3
and for m=2/3
tangent is y=(2/3) x -4/3

Answer 3

hope u understand the method rather than this solution..

Answer 4

is it that u r mastering tangent and normals using derivatives..

Answer 5

the first part i do not understand at all, it got me even more confused but the second part i understand

Answer 6

see what is ur confusion in the first part

Answer 7

i just do not understand how you got 6t, 3t^2 ?

Answer 8

for the parabola x^2=4by
the parametric eq can be written as x=2bt and y=bt^2 t being the parameter
or in other words any point on the above parabola is of the form (2bt,bt^2)

Answer 9

thus here the parabola is x^2=12y or x^2=4(3)y
that is b here is 3 hence any point on the parabola is of the form (2*3*t,3*t^2)

Answer 10

hope u get this
n further say what s still creating confusion

Answer 11

are u there...