Math Is Fun Forum

To futher complete this topic I'll post that I now have a solution for the hyperbola.

The task is:

Given the conic section :

Find the foci.

(If the equation represents a parabola

If the equation represents a hyperbola, which will be the case when

then the method is as follows:

Find the asymptotes of the hyperbola

case B != 0.0:

Gradients given by

Intercepts by

case B = 0.0:

Gradients given by

Intercetps by

This covers all cases of the hyperbola.

Now, find intersection points of these two lines, this is the centre of the hyperbola.

Next find the two lines which bisect the two angles formed by the two asymptotes. Only one of these lines will intersect the hyperbola, the two intersection points are the vertices of the hyperbola.

Let a = the distance from the centre to a vertex.

Next, move from the centre a distance ae along the line joining the two vertices towards each vertex where e is the eccentricity of the hyperbola calculated as:

where n is 1 or -1 as the determinant:

is negative or positive respectively.

This gives the two foci of the hyperbola.

I've got this running nicely in an interactive application and it works very well.

If anyone can think of a more elegant method I'd be very glad to hear.

Onward to the ellipse..........

let

then

and

so

Hence:

We need to show that this is always between -1 and 1 inclusive, that is to say we need to show that the denominator always has greater or equal absolute value than that of the numerator, i.e. that:

i.e. that

i.e. that

i.e. that

i.e. that

Which is clearly true as the LHS is always > 0.

One thing, we have proven the inequality for ALL complex numbers except one, the number 0 + 0i, for this number the inequality is not true as the denominator would be zero.

Quite right. Good fun. The problem is mine.

I write:

And that is true, at any given instant in time for the cone of water.

I next need to calculate

BUT, for the water cone, r is a function of h, in fact r = h/3.

However, I differentiated as if r were a constant wrt h.

I should have substituted in r = h/3 to give:

And THEN differentiated wrt h. Then we of course agree.

Lesson learned (again!).

Never too late,

Thank you. I had eventually solved this one but not this way. It is always instructive and enjoyable to see alternative attacks on given problems.

To prove this we can use the "rearrangement inequality".

This states that given two sets of positive real numbers:

and

and a given rearrangement of X and Y:

The product (and therefore also the sum - as product is just repeated addition):

is maximized when X' and Y' are both ordered in either ascending or descending order.

To prove this consider the case n = 2:

where

and

then

For the general case for any n let

and

such that

and

suppose that

is a permutation of X with

and

is a permutation of Y such that the product

is maximized.

If there exists a pair

then

So interchange of these pairs will increase the product. Well if we interchange all such situations we end up with V and Z ordered in the same way.

So in choosing our numbers we should use a greedy approach:

First choose 9 and 8.

then 7 and 6.

We place the 7 so that the partial products of cross multiplication are maximized so we now have 96 and 87 and so on to give:

9642 * 87531 as the largest product.

Now of course my proof above assumes both sets of numbers have equal cardinality but clearly here we have and odd number of digits and our two subsets are of differing cardinality - but you can fill in the details here, details also which need to be shown to prove that any other splitting of the digits will produce only lesser products e.g. a two digit number times a seven digit number etc etc.

interesting links:

http://en.wikipedia.org/wiki/Inequality … tric_means

and

http://en.wikipedia.org/wiki/Chebyshev%27s_inequality

1) {(0,0),(1,3),(2,1),(3,4),(4,2)}

2) From above the range of f(x) is A. This is because the value of f(x) can take any of all of the values from A = {0,1,2,3,4}.

3)

You can verify this by multiplying each second element in every ordered pair by 2 and reducing modulo 5 and seeing that the answer in each case is the first element in the corresponding ordered pair.

2 is said to be the multiplicative inverse of 3 mod(5)

These formulae are just two different ways of writing the same thing, like x + y and y + x they are just alternative ways of expressing the same quantity.

Notice also that (x-y)^2 is the same as (y-x)^2. If one of x-y and y-x is -'ve so the other will be +'ve, but both will be of the same magnitude. e.g. 7-5=2 and 5-7=-2 etc. But as a -'ve times a -'ve is +'ve we see that (7-5)^2 = 2^2 = (-2)^2 = (5-7)^2

So:

[Removed]

2)

Call this "Equation (1)".

Differentiation gives us the gradient of this curve at any point:

We know that two lines which are mutually perpendicular have gradients whose product is -1. Therefore the gradient of the line we require the equation of is:

This is the gradient of the perpendicular to the curve at any point on the curve with a given x ordinate.

So we know that the equation of the perpendicular to the curve at any given x ordinate will have the form:

Specifically, we are interested in the point where x = a, hence the equation at this point will be:

Call this "Equation (2)".

We just need to know the value for C. Well, we know one point on this line. The point (a,2a^3+7a-3) is on this line - this from equation (1).

So in Equation (2) we simply substitute in X = a and Y = 2a^3+7a-3 to give:

So putting this value of C into Equation (2) gives:

Call this Equation (3).

This is the equation we desire, it now has to be put into general form. At the moment it is in the form y = mx + c but general form would be ax + by + c = 0

So, working on Equation (3) yeilds:

This is our answer.

To see an instance of this result, let's set a=2. Then Equation (2) becomes:

A plot of this should yield a line which cuts the curve perpendicularly at x = 2.

Indeed it does

For low numbers trial and error would be fine, in the example you gave we know that we need to use the formula:

Specifically we need to solve:

for n.

We can simply try integral values of n {1,2,3.....} and we soon arrive at n=8 as our solution.

In general this function will not have an integral solution.

The generalised factorial function is the Gamma function which defines the factorial of real and complex values: http://en.citizendium.org/wiki/Gamma_function

It is the inverse of this function which one would need in general - not a trivial thing. Google "inverse Gamma function" to find out more!