Fool's problem of the day,
\(1.\)Find the locus of the intersection of the perpendicular tangents drawn to the curve \(4y^3=27x^2 \).
Stacey Warren - Expert brainly.com
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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.
If you try to do the easy problem for y = x^2, you find that the locus is the
Since the graph is symmetrical w.r.t. the y-axis, I'll consider first x>0.
By implicit differentiation, we have:
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.