Just another analytic geometry problem,

Find the locus of the point of intersection of the three normals to the parabola $ y^2=4ax$, to of which are inclined at right angles to each other.

[Solved by @Taufique]

PS: Not too hard but it took me about 12 minutes to solve this, I would love to hear your timings.

Question

Just another analytic geometry problem,

Find the locus of the point of intersection of the three normals to the parabola $ y^2=4ax$, to of which are inclined at right angles to each other.

[Solved by @Taufique]

PS: Not too hard but it took me about 12 minutes to solve this,  I would love to hear your timings.

anonymous · Answer

PS: I don't knew that's it's an IIT level problem prior to posting it, so probably too easy for many of you!

anonymous · Answer

hahahaa fool good question just wait answer is coming ...

perl · Answer

|dw:1333448995076:dw|

perl · Answer

find the locus of the three normals? can you define that

perl · Answer

you just want the point where they intersect, three normals to parabola

apoorvk · Answer

i am just thinking that there is a lot more of graphical analysis involved here rather calculations. am i anywhere near being called right? because u seem to have an answer (may be foolishly wrong)

apoorvk · Answer

I mean "I" I seem to have an answer.. "typo"

anonymous · Answer

He wants the point where those 3 normals to the parabola each have a right angle to each others. Bit more tricky then just finding 3 normals that intersect, afaik.

anonymous · Answer

What's afaik?

anonymous · Answer

As far as I know*

anonymous · Answer

Cool!

apoorvk · Answer

so what was a faik?

perl · Answer

is there a book on this? annoying problems with solutions

anonymous · Answer

three normals can be drawn from a point (x1,y1) to the parabola.the points where these normals meet the parabola r callled feet of the normals .the sum of the slopes of these normals is 0.and the sum of the ordinates of the feet of these normals is also 0

anonymous · Answer

How can three normals be perpendicular to each other?

anonymous · Answer

hehhee kyo nhi ho skte

apoorvk · Answer

all three are NOT perpendicular. any two are, @ishaan94. i did the same mistake too while reading the question.

anonymous · Answer

Oh

anonymous · Answer

Nightie ycm

anonymous · Answer

Yeah, 3 cant be :P It'll form a cross in the parabola.

apoorvk · Answer

if all three are perpendicular, then they meet only at (infinty, 0)

anonymous · Answer

hahhaha right apoorvk

perl · Answer

so it looks like a T

perl · Answer

|dw:1333450382013:dw|

perl · Answer

sorry, that is 2 normals

perl · Answer

how can there be three normals at right angles to each other?

anonymous · Answer

The problem I'm facing is the limited knowledge of parabola :( ...and too many variables

anonymous · Answer

-| is how I envision the solution, I have no idea at all how to calculate the point though.

anonymous · Answer

if the three normal are drawn from a common point (h,k) then their slopes are the roots of equation.
$$K =mh-2am-am ^{3}$$
|dw:1333450272122:dw|
since the slopes are the roots of this equation then
$$m1\times m2 \times m3=-k \div a$$
since $$m1\times m2=-1$$
therefore $$m3=k \div a$$
since the equation of any normal to the parabola Y^2=4aX be 
$$Y =mX-2am-am ^{3}$$.......since this equation passes through (h.k) then put this value in this equation and put m=m3.
after putting this value in this equation replace h and k by X and Y respectively this gives locus of the point of intersection of these three normals........