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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
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This is somewhat difficult, took me some time to realize the solution.
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.
Cheater. ;P

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Other answers:

y=3 ???
No.
Darn I need to read up on English definitions.ъ :(
man, this is hard questions!!!
@FoolForMath , what do you mean by perp tangent/
is it x=0?
Sorry, that's not the right answer.
teehee :)
When you say perpendicular tangents, do you mean two tangents lines of the curve that are perpendicular to each other?
http://www.wolframalpha.com/input/?i=plot+4y^3%3D27x^2
If you try to do the easy problem for y = x^2, you find that the locus is the line y=-1/4.
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?
Why do you Google Cinar? I am well aware of the answer.
I am removing the spoiler.
@blockcolder: looks good.
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
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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