1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parabola Three Normals

  1. Sep 16, 2012 #1

    utkarshakash

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Three normals are drawn from the point (14,7) to the curve [itex]\large y^{2} -16x-8y=0[/itex]. Find the co-ordinates of the feet of the normals.

    2. Relevant equations
    Converting the equation of parabola in the form of a perfect square I get
    [itex]\large (y-4)^{2}=16(x+1)[/itex]

    3. The attempt at a solution
    I know that for the parabola [itex]\large y^{2}=4ax[/itex] the equation of normal is [itex]\large y=-tx+2at+at^{3}[/itex]. But what will be the equation of normal for the parabola in this question? If I get to know the equation of normal in terms of t then I can get a cubic equation in t and by solving that equation I can obtain the coordinates of feet of normals. But I'm falling short of the equation of normal. How do I get it?
     
  2. jcsd
  3. Sep 16, 2012 #2

    Mentallic

    User Avatar
    Homework Helper

    Remember how if we have some parabola y=x2 and we then substitute x'-1 for x, to get y=(x'-1)2, this is now a parabola that has moved 1 unit to the right. Also, if we substitute y'-1 for y and get y'-1=(x'-1)2 this is now a parabola that now moved up 1 unit as well, so its apex is now at (1,1).

    This also applies to all functions. So if we have a tangent to the parabola y=x2 as y= mx+c, then if we move the parabola across a units and up b units (a and b could be negative too so we can move then down and left) to get the parabola y-b=(x-a)2 then if we do the same transformation to y=mx+c to get y-b=m(x-a)+c this will now still be a tangent to the parabola at the same point.

    So where am I going with this?

    Well, you've already told me that

    So then what is the normal for the parabola [itex]\large (y-4)^{2}=16(x+1)[/itex] now?
     
  4. Sep 16, 2012 #3

    utkarshakash

    User Avatar
    Gold Member

    Yep. That worked! Thank You once again for helping me. Can you please solve my another question which I have posted in this forum titled "Locus of circumcentre". I really need some hints to get started with.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Parabola Three Normals
  1. Tangents and Parabolas (Replies: 3)

  2. Normal to parabola (Replies: 3)

Loading...