anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
hahahaa fool good question just wait answer is coming ...
perl
  • perl
|dw:1333448995076:dw|

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perl
  • perl
find the locus of the three normals? can you define that
perl
  • perl
you just want the point where they intersect, three normals to parabola
apoorvk
  • 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
  • apoorvk
I mean "I" I seem to have an answer.. "typo"
anonymous
  • 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
  • anonymous
What's afaik?
anonymous
  • anonymous
As far as I know*
anonymous
  • anonymous
Cool!
apoorvk
  • apoorvk
so what was a faik?
perl
  • perl
is there a book on this? annoying problems with solutions
anonymous
  • 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
  • anonymous
How can three normals be perpendicular to each other?
anonymous
  • anonymous
hehhee kyo nhi ho skte
apoorvk
  • apoorvk
all three are NOT perpendicular. any two are, @ishaan94. i did the same mistake too while reading the question.
anonymous
  • anonymous
Oh
anonymous
  • anonymous
Nightie ycm
anonymous
  • anonymous
Yeah, 3 cant be :P It'll form a cross in the parabola.
apoorvk
  • apoorvk
if all three are perpendicular, then they meet only at (infinty, 0)
anonymous
  • anonymous
hahhaha right apoorvk
perl
  • perl
so it looks like a T
perl
  • perl
|dw:1333450382013:dw|
perl
  • perl
sorry, that is 2 normals
perl
  • perl
how can there be three normals at right angles to each other?
anonymous
  • anonymous
The problem I'm facing is the limited knowledge of parabola :( ...and too many variables
anonymous
  • anonymous
-| is how I envision the solution, I have no idea at all how to calculate the point though.
anonymous
  • 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
  • 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
  • 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
  • perl
i dont understand Taufique's solution
anonymous
  • anonymous
Me neither :(
anonymous
  • anonymous
I am not getting the right approach, too many variables
karatechopper
  • karatechopper
i gotta get my mints in hand if i want to solve this! sounds easy! WITH MINTS! ;D
anonymous
  • 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
  • apoorvk
mints? you need RED-BULL for this one.
karatechopper
  • karatechopper
but he said mints!
anonymous
  • anonymous
the line of symmetry for the parabola is the locus
anonymous
  • anonymous
He? I thought you were she karate
karatechopper
  • karatechopper
the lotus flower? line of symmetry is lotus!! jkjkjk!
karatechopper
  • karatechopper
i know u said locus:)
anonymous
  • 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
  • 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
  • anonymous
That link has nothing to do with the problem, probably. (I didn't read it)
karatechopper
  • karatechopper
i didnt read it
apoorvk
  • 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
  • karatechopper
is the answer a curved line? hmm lemme try
apoorvk
  • apoorvk
all i can spawn is that the locus is another parabola, with its vertex probably at the focus of the first one.
apoorvk
  • apoorvk
yeah yeah KC, the answer is definitely some curved line. BUT WHAT IS ITS EQUATION????? :/
.Sam.
  • .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.
  • .Sam.
@Ishaan94 any ideas from here?
anonymous
  • anonymous
Unfortunately No, that's where I hit the roadblock.
apoorvk
  • 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
  • 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
  • anonymous
So close ...yet so far. I almost had it :(
apoorvk
  • 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
  • apoorvk
ok.. you missed the sgrt over there i guess.
anonymous
  • 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
  • apoorvk
'cause i solved in terms of y. no problem
anonymous
  • anonymous
FoolForMath Y U NO POST THE SOLUTION?
anonymous
  • anonymous
Nightie you still on it?
anonymous
  • anonymous
I did this one exactly in the same way as @Taufique

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