I am going to post 1 math problem a week. (If that's ok with Yoni) They shall vary in diffaculty and what field they are in.

First one is a tricky, diffacult one.

*I make no claim I know the awnser or even to have made the problem up, they are just posted as is, for you to figure out, for fun.

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Find the equation of the line tangent to the ellipse
b^2*x^2 + a^2*y^2 = a^2*b^2
in the first quadrant that forms with the coordinate axes
the triangle of smallest possible area
(a & b are positive constants)

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Just FYI, it doesn't matter (in this particular problem) whether or not a or b are positive. They're squared in every instance. :P

Where did you get this, anyway? It occurs to me that given a = b = 1, this problem would be relatively easy, but it also occurs to me that it's LIKELY that you could have smaller constants work for a and b (although I haven't had calc for a couple years). Which means, using a circle, you can have infinitessimally small triangles.

I'm sorry but that is not correct.
I cannot give you the source of this problem, for it has the solution also.

[quote author=Snake link=board=36;threadid=8506;start=0#msg78515 date=1094016167]
I am going to post 1 math problem a week. (If that's ok with Yoni) They shall vary in diffaculty and what field they are in.
[/quote]
Of course that's ok. Enjoy :)

Interesting problem. Looks somewhat highschool-math-ish though.
I'll try to solve it when I get back home and PM you the answer.

MyndFyre: It's true that the sign of a,b doesn't matter, but with ellipses (and hyperbolas) it's usually taken as positive due to their meaning (intersection with positive x and positive y axis in the case of an ellipse).

*Tumble weed blows by..

Since no one has figured this problem out, and it has been 1 week, here is the awnser from Doctor.Math:

[quote]First, we'll try to find the equation of a line that's tangent to the
ellipse. Since you said that we're looking for tangency in the first
quadrant, I'll write the equation of the ellipse as

y = b*Sqrt{a^2 - x^2}/a

I've just solved for y here. Then to find the slope of the tangent line
to the ellipse, I'll take the derivative of this expression:

y' = -bx/(a*Sqrt{a^2 - x^2})

So if we let r be some value of x that gives us a point on the ellipse in
the first quadrant, the equation of the line tangent to the ellipse at
(r,b*Sqrt{1-(1/a)^2*r^2} will have the form

y = (-br/(a*Sqrt{a^2 - r^2})) x + c.

The next step is to find c in terms of a, b, and r. So let's do it. The
usual way is to plug in (for x and y) some point that you know is on the
line, and in this case we'll pick the point (r, b*Sqrt{a^2 - r^2}/a). So
we get

b*Sqrt{a^2 - r^2}/a = -br^2/(a*Sqrt{a^2 - r^2}) + c


c = b*Sqrt{a^2 - r^2}/a + br^2/(a*Sqrt{a^2 - r^2})

b(a^2 - r^2) b*r^2 ab
c = ----------------- + ----------------- = ---------------
a*Sqrt{a^2 - r^2} a*Sqrt{a^2 - r^2} Sqrt{a^2 - r^2}


So the equation for the line tangent to the ellipse at the point
(r,b*Sqrt{a^2-r^2}/a) is

-brx ab a^2*b - brx
y = ----------------- + --------------- = -----------------
a*Sqrt{a^2 - r^2} Sqrt{a^2 - r^2} a*Sqrt{a^2 - r^2}


Now take a breath. We've completed the first part of the problem.
Go get a glass of orange juice or something, because there's more,
and we're going to start to use Calculus.

Okay, we're back. We want to find out the area of the region bounded by
this line and the x- and y- axes. Well, it's a right triangle, so we'll
use the equation Area = Base x Height / 2. The base and the height are
going to be, respectively, the x-intercept and the y-intercept of the
line we found. So let's find the intercepts:

To find the y-intercept, we plug in x = 0 and solve for y, finding
y = ab/Sqrt{a^2 - r^2}. Similarly, to find the x-intercept, we plug in
y = 0 and solve for x, finding that x = a^2 / r. So the area of the
triangle is going to be the product of these numbers divided by 2, i.e.
a^3*b/(2r*Sqrt{a^2 - r^2}.


Take another breath. A smaller breath, though, because that part wasn't
as hard. Now we need to treat this like an ordinary Max/Min problem,
and find the minimum value of this Area function. The way we do this
is to take its derivative, and then set that equal to zero to find
critical points. So let's do it. Remember, the area is a function of
only one variable, since a and b are fixed; they're given to us in the
beginning of the problem.

-a^3*b (Sqrt{a^2 - r^2} - r^2 / Sqrt{a^2 - r^2})
Area' = ------------------------------------------------
2r^2 (a^2 - r^2)

Check my work here. Not only is it important that you see how the
computations work, I might have messed up.

So we want to set this equal to zero and solve for r. Remember that a
fraction is zero when its numerator is zero, so we solve the equation:

-a^3*b(Sqrt{a^2 - r^2} - r^2/Sqrt{a^2 - r^2}) = 0.

a^5*b
I get r^2 = ---------
a^3*b - 1[/quote]

New problem to be posted.