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#1 2012-01-04 22:28:45

juantheron
Member
Registered: 2011-10-19
Posts: 312

parabola

Let

be the latusrectum of the parabola
with vertex

Then Minimum Length of Projection of

on a tangent drawn in the portion
, is

answer =

actually i did not understand the question

can anyone draw the figure of that problem

thanks

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#2 2012-01-05 01:55:05

Bob
Administrator
Registered: 2010-06-20
Posts: 10,058

Re: parabola

hi juantheron,

Here's a diagram. 

I'm unclear what is meant by the 'projection'.

Bob

Last edited by Bob (2012-01-05 01:56:13)


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#3 2012-01-05 12:08:21

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: parabola

I'm not completely sure but I think this is what they mean by "projection on the tangent".  It gives an answer of 2√2 at least.


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#4 2012-01-06 01:00:55

juantheron
Member
Registered: 2011-10-19
Posts: 312

Re: parabola

The Dude plz  can you me how can you get it

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#5 2012-01-08 03:52:33

TheDude
Member
Registered: 2007-10-23
Posts: 361

Re: parabola

The short answer is "one step at a time".  This is the list of things we need to do for this problem:

1. Find a formula for the equation of the tangent line at a given point on the parabola.
2. Find the points on this tangent line that form the endpoints of the projection of the latus rectum.
3. Find a formula for the distance between these two points.
4. Find the minimum possible distance over the region we are told to work with.

Start by noting that the parabola is symmetrical so we only need to consider half of it, we'll go with the top half.  This allows us to simplify the equation from y²=4x to y=2√x.  Taking the derivative gives us y'=1/√x.  So for a given x-coordinate 'c', the slope of the tangent line at (c, 2√c) is 1/√c.  This makes the full equation of the tangent line y= (1/√c)x + b.  To solve for 'b' we plug in our point: 2√c = (1/√c)c + b, b = 2√c - c/√c, b = c/√c.  So our tangent line is y= (1/√c)x + √c.

For step 2 we need to find the endpoints of the projection of the latus rectum onto the tangent line.  The endpoints of the latus rectum itself are (1, 2) and (1, -2).  If you aren't sure why this is, the article on Wikipedia can explain.  To find the projection of these points we need the lines that pass through these points and are perpendicular with the tangent line, which means their slopes will be -√c.  For the point (1, 2) we have 2= (1)(-√c) + b, b=2 + √c, so we need to find the intersection between y= (1/√c)x + √c and y=-x√c + 2 + √c.  Set them equal to get (1/√c)x + √c = -x√c + 2 + √c, x(√c + 1/√c) = 2, x = 2/(√c + 1/√c) = 2/( [c + 1] / √c) = (2√c) / (c + 1).  Solving for y gives us y = (2 + c√c + √c) / (c + 1), which finally gives us ( [2√c]/[c + 1], [2 + c√c + √c]/[c + 1] ) as the first endpoint of the projection.  Doing the same for (1, -2) gives us ( [-2√c]/[c + 1], [-2 + c√c + √c]/[c + 1] ).

For step 3 we us the standard formula for the distance between two points, which is

This actually turns out pretty well for us:

To recap, what we have so far is the formula for the length of the projection of the latus rectum onto the tangent line at the point (c, 2√c).  Now we want to find the minimum possible value over the region c ∈ [0, 1].  Taking the derivative gives us

We can see that there are no critical points in the region we are working with, so we just need to evaluate the endpoints, or c=0 and c=1.  Solving for those gives us 4/√1 = 4, and 4/√2 = 2√2.  So our final answer is that the minimum length of the projection is 2√2 and occurs at the point (1, 2).  Since the parabola is symmetrical it also occurs at (1, -2).

Last edited by TheDude (2012-01-08 03:54:25)


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#6 2012-01-08 05:25:50

juantheron
Member
Registered: 2011-10-19
Posts: 312

Re: parabola

Thanks Dude

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