# Quick question-Representing a curve w/a DE

1. Sep 9, 2008

### bcjochim07

Assume that when the plane curve C shown in the figure is revolved about the x axis, it generates a surface with the property that all light rays L parallel to the x axis striking the surface at a point P(x,y) are reflected to a single point O (the origin). Use the fact that the angle of incidence is equal to the angle of reflection to determine a differential equation that describes the shape of the curve C.

My work:

The tangent of angle that the tangent line makes with the x axis is the slope of that tangent line, dy/dx. Using some simple trig, I was able to determine that angle phi = 2theta, so its supplementary angle is 180-2theta. Therefore the angle that the tangent makes with the x axis is theta. The distance between the origin is sqrt(x^2+y^2), and this is an isoceles triangle, so the distance between the origin and where the tangent line intersects the x axis is also sqrt(x^2+y^2). Drawing a right triangle to find out what the tangent of theta is, I have a triangle with legs length sqrt(x^2+y^2)-x and y
So I came up with tan theta = dy/dx = y/(sqrt(x^2+y^2)-x). Why does my book say that the answer is (sqrt(x^2+y^2)-x)/y?

2. Sep 11, 2008

### bcjochim07

Help would be appreciated. Thanks

3. Sep 11, 2008

### matematikawan

You have chosen the second quadrant as your general point so it must be P(-x,y) not P(x,y) which should be located in the first quadrant.

But never mind, we just stick with what you already have. So changing all your x to -x, your right angled triangle should have the sides (x^2+y^2)+x and y.

$$tan\theta=\frac{dy}{dx}= \frac{y}{\sqrt{x^2+y^2}+x} = \frac{\sqrt{x^2+y^2}-x}{y}$$

4. Sep 13, 2008

### bcjochim07

Thanks.

There is another way to do this using the fact that phi is equal to 2* theta & using the identity tan (x/2) = +/- sqrt [(1-cosx)/(1+cosx)] I am having trouble figuring out and expression for the cosine of the angle phi, since I cannot draw in a right triangle.
I have also tried to use the law of sines and law cosines, but I'm not having much luck. Any pointers about how to use this identity?

5. Sep 14, 2008

### bcjochim07

any ideas?

6. Sep 14, 2008

### bcjochim07

Nevermind, I figured it out.