# Orthogonal polynomials

1. Jul 31, 2015

### Bruno Tolentino

Given a quadric equation (F(x,y) = 0), exist other quadric equation (G(x,y) = 0) such that the poinst of intersection between the graphics are ortogonals.

So, how to find the coefficients of the new quadic equation?

EDIT: I think that F and G needs to satisfyF • G = 0. So, if F is known, how discover G?

Last edited: Jul 31, 2015
2. Jul 31, 2015

### Dr. Courtney

How can points be orthogonal?

The intersections of two surfaces are most commonly curves.

I am not sure what is being asked.

3. Aug 1, 2015

### Bruno Tolentino

The crossings are orthogonals.

4. Aug 1, 2015

### HallsofIvy

You mean "the curves are orthogonal at every point of intersection".

5. Aug 2, 2015

### Bruno Tolentino

Yeah! So, exist some way of find the G function?

6. Aug 7, 2015

### HallsofIvy

First, in order to have what you are talking about, you have to have, not a single function, F(x, y), but a family of functions.

To take a simple example, look at $x^2+ y^2= R^2$, the family of all circles with center at the origin. Differentiating both sides of the equation with respect to x, $2x+ 2y dy/dx= 0$ so that $dy/dx= -x/y$ at every point. To be orthogonal to that, a function, y(x), must have derivative equal to the negative of the reciprocal. That is, we want $dy/dx= -1/(-x/y)= y/x$. That is, the "orthogonal complement" of this family of curves must satisfy $dy/dx= y/x$.
That is an easily separable equation- it can be written $dy/y= dx/x$. Integrating both sides, $ln(|y|)= ln(|x|)+ C$ so that $|y|= C'|x|$ where C' is equal to the exponential of C. By allowing C to take on non-positive values also, we can drop the absolute values and have $y= Cx$. That is the family of all straight lines that go through the origin. They are diameters of the original circles so always perpendicular to them.

More generally, given a family of curves, functions of x and y that depend a constant, to find the orthogonal complement, differentiate the equation defining the family to eliminate that constant. Find dy/dx= m(x,y) from that equation and then solve the differential equation dy/dx= -1/m(x,y).

7. Aug 8, 2015