anonymous
  • anonymous
Fool's problem of the day, \(1.\)Find the locus of the intersection of the perpendicular tangents drawn to the curve \(4y^3=27x^2 \). Good luck!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
This is somewhat difficult, took me some time to realize the solution.
anonymous
  • anonymous
Where does one start to find the answer to this question ? you can msg me so you don't give it away if you want, I just wanna know HOW to solve it.
anonymous
  • anonymous
Cheater. ;P

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experimentX
  • experimentX
y=3 ???
anonymous
  • anonymous
No.
anonymous
  • anonymous
Darn I need to read up on English definitions.ъ :(
anonymous
  • anonymous
man, this is hard questions!!!
inkyvoyd
  • inkyvoyd
@FoolForMath , what do you mean by perp tangent/
inkyvoyd
  • inkyvoyd
@FoolForMath ?
asnaseer
  • asnaseer
is it x=0?
anonymous
  • anonymous
Sorry, that's not the right answer.
anonymous
  • anonymous
teehee :)
blockcolder
  • blockcolder
When you say perpendicular tangents, do you mean two tangents lines of the curve that are perpendicular to each other?
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=plot+4y^3%3D27x^2
anonymous
  • anonymous
Yes, @blockcolder
anonymous
  • anonymous
If you try to do the easy problem for y = x^2, you find that the locus is the line y=-1/4.
blockcolder
  • blockcolder
Since the graph is symmetrical w.r.t. the y-axis, I'll consider first x>0. By implicit differentiation, we have: \[4y^3=27x^2\\ 12y^2y'=54x\\ y'=\frac{54x}{12y^2}=\frac{9x}{2y^2}\] Am I on the right track?
anonymous
  • anonymous
Why do you Google Cinar? I am well aware of the answer.
anonymous
  • anonymous
I am removing the spoiler.
anonymous
  • anonymous
@blockcolder: looks good.
dumbcow
  • dumbcow
the set of points that represent the intercepts of the perpendicular tangents match the original function. all the intercepts lie on the same curve it was a little bit of work so i won't post my solution here...the steps were 1) find dy/dx, then the perpendicular slope is -dx/dy 2) solve for line equations for each slope 3) set them equal to find intercept points
asnaseer
  • asnaseer
ok - I misunderstood the question. I took it to mean the intersection of the lines that are perpendicular to the tangents to the given curve.

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