## anonymous 4 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!

1. anonymous

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

2. anonymous

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.

3. anonymous

Cheater. ;P

4. experimentX

y=3 ???

5. anonymous

No.

6. anonymous

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

7. anonymous

man, this is hard questions!!!

8. inkyvoyd

@FoolForMath , what do you mean by perp tangent/

9. inkyvoyd

@FoolForMath ?

10. asnaseer

is it x=0?

11. anonymous

Sorry, that's not the right answer.

12. anonymous

teehee :)

13. blockcolder

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

14. anonymous
15. anonymous

Yes, @blockcolder

16. anonymous

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

17. blockcolder

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?

18. anonymous

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

19. anonymous

I am removing the spoiler.

20. anonymous

@blockcolder: looks good.

21. dumbcow

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

22. asnaseer

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.