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

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- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

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

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

- perl

|dw:1333448995076:dw|

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## More answers

- perl

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

- perl

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

- apoorvk

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

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

- anonymous

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

What's afaik?

- anonymous

As far as I know*

- anonymous

Cool!

- apoorvk

so what was a faik?

- perl

is there a book on this? annoying problems with solutions

- anonymous

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

How can three normals be perpendicular to each other?

- anonymous

hehhee kyo nhi ho skte

- apoorvk

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

- anonymous

Oh

- anonymous

Nightie ycm

- anonymous

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

- apoorvk

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

- anonymous

hahhaha right apoorvk

- perl

so it looks like a T

- perl

|dw:1333450382013:dw|

- perl

sorry, that is 2 normals

- perl

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

- anonymous

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

- anonymous

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

- anonymous

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........

- apoorvk

i have a very long equation....
and i think its totally wrong, yet i ll post it.
\[y^6 - 4a^2xy^4 +8a^2y^4 -16xy^2+32a^4y^2+64a^2 = 0\]

- apoorvk

since they intersect, i equated the two normals, and since they are perpendicular, slope of any one normal is negative of reciprocal of the other's.

- perl

i dont understand Taufique's solution

- anonymous

Me neither :(

- anonymous

I am not getting the right approach, too many variables

- karatechopper

i gotta get my mints in hand if i want to solve this! sounds easy! WITH MINTS! ;D

- anonymous

Okay, first FoolForMath labels it as silly and now an 8th grader tells me this is easy...
Something is really wrong with me :/

- apoorvk

mints? you need RED-BULL for this one.

- karatechopper

but he said mints!

- anonymous

the line of symmetry for the parabola is the locus

- anonymous

He? I thought you were she karate

- karatechopper

the lotus flower? line of symmetry is lotus!! jkjkjk!

- karatechopper

i know u said locus:)

- anonymous

No, line of symmetry can't be its locus, if it's then I should just jump off the cliff. Only if I had a cliff nearby hehe

- anonymous

go through this !! -http://www.jstor.org/discover/10.2307/2300595?uid=3738256&uid=2129&uid=2&uid=70&uid=4&sid=21100699783861

- anonymous

That link has nothing to do with the problem, probably. (I didn't read it)

- karatechopper

i didnt read it

- apoorvk

It can't be the line of symmetry!! check this out.
|dw:1333452500834:dw|
and ishaan, how come i think the same things you think? i am feeling like a stoooopid!!

- karatechopper

is the answer a curved line? hmm lemme try

- apoorvk

all i can spawn is that the locus is another parabola, with its vertex probably at the focus of the first one.

- apoorvk

yeah yeah KC, the answer is definitely some curved line. BUT WHAT IS ITS EQUATION????? :/

- .Sam.

If the parabola look something like this,
|dw:1333453961998:dw|
then the gradient of normals will be 0
-------------------------------------------------------------
If it looked something like this|dw:1333454121058:dw|
There's only 1 normal.
-------------------------------------------------------------
If it looked something like this |dw:1333454716267:dw|
The lines are parallel each other,
So,
\[y^2=4ax\]
y=(4ax)^{1/2}
\[dy/dx=\frac{1}{2}(4ax)^{-1/2}(4a)....................(Gradient ~1)\]
m1m2=-1
\[m_{2}=-\frac{\sqrt{4ax}}{2a}................................(Gradient~2)\]
--------------------------------------------------------------
From here, use y-y1=m(x-x1) for both gradients and equate each other.
\[2(a)(4ax)^{-1/2}(x-x_{1})+y_{1}=-\frac{(\sqrt{4ax})}{2a} (x-x_{1})+y_{1}\]
Simplifying you get
x=-a then you need to find 'y'

- .Sam.

@Ishaan94 any ideas from here?

- anonymous

Unfortunately No, that's where I hit the roadblock.

- apoorvk

the problem is you get parallel normals only at infinity, when the curves of the the parabola are retricemptotically parallel to its axis. so what you drew sam, is a bit exaggerated it seems to me?

- anonymous

Let \((h,k)\) be the point of intersection, and \((x_1,y_1)\), \((x_2,y_2)\) be the point of contact for the two perpendicular Normals to the curve \(y^2 = 4ax\).\[\frac{y_1-k}{x_1-h}=-\frac{x_2-h}{y_2-k}\]\[\implies y_2y_1 + k^2 -ky_2 - ky_1=-x_1x_2 -h^2 + hx_1+kx_2\]\[\implies y_2y_1+x_1x_2+k^2+h^2=h(x_1+x_2)+k(y_1+y_2)\tag1\]Equation of the curve is \(y^2 = 4ax\)\[\implies \frac{dy}{dx} = \frac{a}{x}\]Because the two Normals are perpendicular,\[\sqrt \frac ax_1 = -\sqrt\frac{x_2}{a}\implies -a = \sqrt{x_1x_2}\implies x_1x_2 = a^2\]Similarly, \(y_1y_2=-4a^2\).
Using equation (1),\[\implies -4a^2 + a^2+k^2 +h^2 = h(x_1 +x_2)+k(y_1+y_2)\]
I need one more equation or, maybe I overlooked something.

- anonymous

So close ...yet so far. I almost had it :(

- apoorvk

real good attempt. i got that far, and stuck too. but, one thing ishaan, wouldn't dy/dx there be (2a/y) instead of (a/x)? or am i dreaming again?

- apoorvk

ok.. you missed the sgrt over there i guess.

- anonymous

I can express \(x_1 + x_2\) as \(\sqrt{(x_1-x_2)^2 + 4x_1x_2}\). And maybe then I could use some relation, perhaps pythagorus theorem? No, that wouldn't help either.
Yeah, my bad I missed the square root.

- apoorvk

'cause i solved in terms of y. no problem

- anonymous

FoolForMath Y U NO POST THE SOLUTION?

- anonymous

Nightie you still on it?

- anonymous

I did this one exactly in the same way as @Taufique

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