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Parabola Three Normals

  • #1
utkarshakash
Gold Member
855
13

Homework Statement


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.

Homework Equations


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

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?
 

Answers and Replies

  • #2
Mentallic
Homework Helper
3,798
94

Homework Statement


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.

Homework Equations


Converting the equation of parabola in the form of a perfect square I get


The Attempt at a Solution


. 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?
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

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]
So then what is the normal for the parabola [itex]\large (y-4)^{2}=16(x+1)[/itex] now?
 
  • #3
utkarshakash
Gold Member
855
13
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.
 

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