Question

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!

Answer 1

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

Answer 2

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.

Answer 3

Cheater. ;P

Answer 4

y=3 ???

Answer 5

No.

Answer 6

Darn I need to read up on English definitions.ъ :(

Answer 7

man, this is hard questions!!!

Answer 8

@FoolForMath , what do you mean by perp tangent/

Answer 9

@FoolForMath ?

Answer 10

is it x=0?

Answer 11

Sorry, that's not the right answer.

Answer 12

teehee :)

Answer 13

When you say perpendicular tangents, do you mean two tangents lines of the curve that are perpendicular to each other?

Answer 14

http://www.wolframalpha.com/input/?i=plot+4y^3%3D27x^2

Answer 15

Yes, @blockcolder

Answer 16

If you try to do the easy problem for y = x^2, you find that the locus is the
line y=-1/4.

Answer 17

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?

Answer 18

Why do you Google Cinar? I am well aware of the answer.

Answer 19

I am removing the spoiler.

Answer 20

@blockcolder: looks good.

Answer 21

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

Answer 22

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.