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FoolForMath
 3 years 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!
FoolForMath
 3 years 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!

This Question is Closed

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1This is somewhat difficult, took me some time to realize the solution.

rebeccaskell94
 3 years ago
Best ResponseYou've already chosen the best response.0Where 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.

prealgebra
 3 years ago
Best ResponseYou've already chosen the best response.0Darn I need to read up on English definitions.ъ :(

ZhangYan
 3 years ago
Best ResponseYou've already chosen the best response.0man, this is hard questions!!!

inkyvoyd
 3 years ago
Best ResponseYou've already chosen the best response.0@FoolForMath , what do you mean by perp tangent/

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1Sorry, that's not the right answer.

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

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

blockcolder
 3 years ago
Best ResponseYou've already chosen the best response.0Since 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?

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1Why do you Google Cinar? I am well aware of the answer.

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1I am removing the spoiler.

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1@blockcolder: looks good.

dumbcow
 3 years ago
Best ResponseYou've already chosen the best response.0the 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
 3 years ago
Best ResponseYou've already chosen the best response.0ok  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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