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Fruityloop

Member

Registered: 2009-05-18

Posts: 143

Tough calculus problem..

I'm stuck on the following Calculus problem..

A tangent is drawn to the ellipse

so that the part intercepted by the coordinate axes is a minimum.
Show that the length is 9 units.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Tough calculus problem..

Hi Fruityloop;

I don't understand what will be nine units?

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

Fruityloop

Member

Registered: 2009-05-18

Posts: 143

Re: Tough calculus problem..

The 9 units is the length from coordinate axis to coordinate axis I believe.
This is apparently the minimum length for all of the possible tangents to the ellipse.
I don't know how to prove this.
I can get the slope of the tangent to the ellipse but then I'm stuck.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Tough calculus problem..

so that the part intercepted by the coordinate axes is a minimum.

My interpretation of this is:

Let T be line tangent to the ellipse.  Set a to be the x-intercept and b to be the y-intercept of T.  Now set d to be the distance between a and b, or infinity if one of the intercepts does not exist.

Minimize d.

Is that right?

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

Fruityloop

Member

Registered: 2009-05-18

Posts: 143

Re: Tough calculus problem..

OK! I finally got the answer!
The key is to make a little rectangle underneath the ellipse, where the upper right corner touches the ellipse at the point of tangency.
In this case, I called the part from the rectangle up to where the tangent intercepts the y-axis, h.
I called the part from the rectangle to where the tangent intercepts the x-axis, b.
The slope of the tangent to the ellipse is

Both the h divided by the x-axis part of the rectangle and the y-axis part of the rectangle divided by b are equal to the slope of the tangent (the absolute value).
So we have

and we have

So the square root of the following gives the distance...

which simplifies to..

Now the length of the tangent between the coordinate axes is the square root of this, but if we take the derivative of this instead it should still give us the right answer.
So taking the derivative and setting it equal to zero and solving for x (a bit of work here)
we end up with

and

as the point on the ellipse where the minimum tangent touches it.
We already know the slope so we can get the equation of the line and find out the length by the pythagorean theorem.
The x-axis distance is

and the y-axis distance is 6.
So squaring both and adding and taking the square root gives 9.

Last edited by Fruityloop (2010-01-07 01:35:38)

Zac

Guest

Re: Tough calculus problem..

umm i see where you are coming from, but i could not follow your working, i have got the same question and it is giving me all kinds of headaches, can you possibly explain it alittle clearer?
Thanks

soroban

Member

Registered: 2007-03-09

Posts: 452

Re: Tough calculus problem..

. .

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ASHOK kUMAR

Guest

Re: Tough calculus problem..

This is not that tough but it is a good problem. Here goes:

The tangent on point (a, b) of the ellipse is: xa/25 +yb/16 =1 which intercepts axes at (0, 16/b) and (25/a, 0).
Hence the intercept between axes of this tangent is: (25/a)² + (16/b)²

Therefore the problem is:  Minimize √ ((25/a)² + (16/b)²) subject to (a/5)² + (b/4)² = 1.
Let's substitute (a/5)² = c and (b/4)² = d.

Then the problem is:  Minimize  ((25/c) + (16/d))² subject to c + d = 1.

Hence the problem is:  Minimize 25/c + 16/(1-c) (dropping the square).

Differentiating wrt  c and placing = 0, we get: 5/ c = 4/(1-c) so that c = 5/9 and d = 4/9.

Therefore, intercept = √ (25/c + 16/(d)) =  = √(45 + 36) = √(81) = 9.

Also, the point is: a = 5√5/3, b = 4√4/3 reversing the substitution for c and d.