cambridge book qn (1 Viewer)

[bos1234]

Show that the eqn of tghe normal to the curve x^2 = 4y at the pt (2t,t^2) is
x + ty -2 t - t^3 = 0

 

[fishy89sg]

fine ill reply

Show that the eqn of tghe normal to the curve x^2 = 4y at the pt (2t,t^2) is
x + ty -2 t - t^3 = 0


x^2 = 4y
y = x^2 / 4
dy/dx = x/2 (gradient function of tangent)
dx/dy = 2/x (gradient function of normal)
at (2t, t^2)
dx/dy = 2/2t = 1/t

Using the point gradient formula

y - (t^2) = 1/t (x - 2t)
ty - t^3 = x - 2t
ty - t^3 - x + 2t = 0

.. why are my + and - off. anyways u get the idea :O

 

[ianc]

you stuffed up the bit where you work out the gradient of the normal from the gradient of the tangent.

m_T x m_N = -1
m_N = -1/m_T

Heres the whole solution:

This is a very standard sort of question for the locus/parabola topic and you will need to get fast at these, so keep practicing.

 

[bos1234]

yes ok thanks..
it confused me because this qn is under factor theorem and remainder theorem..

should have know anywaythanks goodnight

and how did u get the pictures on this site.. i upload to imageshack then copy.. but it doesnt show the picture. only text

 

[ianc]

what you do is upload to imageshack, then type:

[noparse]

[/noparse] (where url is the direct link to the image)

 

[bos1234]

kk :wave:

 

[fishy89sg]

ewwww my stupid mistake