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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!
 one year ago
 one year ago
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!
 one year ago
 one year ago

This Question is Closed

FoolForMathBest ResponseYou've already chosen the best response.1
This is somewhat difficult, took me some time to realize the solution.
 one year ago

rebeccaskell94Best ResponseYou've already chosen the best response.0
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.
 one year ago

prealgebraBest ResponseYou've already chosen the best response.0
Darn I need to read up on English definitions.ъ :(
 one year ago

ZhangYanBest ResponseYou've already chosen the best response.0
man, this is hard questions!!!
 one year ago

inkyvoydBest ResponseYou've already chosen the best response.0
@FoolForMath , what do you mean by perp tangent/
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Sorry, that's not the right answer.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
When you say perpendicular tangents, do you mean two tangents lines of the curve that are perpendicular to each other?
 one year ago

cinarBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=plot+4y^3%3D27x^2
 one year ago

eliassaabBest ResponseYou've already chosen the best response.0
If you try to do the easy problem for y = x^2, you find that the locus is the line y=1/4.
 one year ago

blockcolderBest ResponseYou've already chosen the best response.0
Since the graph is symmetrical w.r.t. the yaxis, 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?
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Why do you Google Cinar? I am well aware of the answer.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
I am removing the spoiler.
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
@blockcolder: looks good.
 one year ago

dumbcowBest ResponseYou've already chosen the best response.0
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
 one year ago

asnaseerBest ResponseYou've already chosen the best response.0
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.
 one year ago
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