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FoolForMath
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 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.
Darn I need to read up on English definitions.ъ :(
man, this is hard questions!!!
@FoolForMath , what do you mean by perp tangent/
Sorry, that's not the right answer.
When you say perpendicular tangents, do you mean two tangents lines of the curve that are perpendicular to each other?
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.