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 December 11th, 2013, 02:38 PM #1 Newbie   Joined: Nov 2013 Posts: 18 Thanks: 0 Calculus of Parabola Term Question So Ive been studying reflective properties of light and so forth and have been trying to fully understand what each term is when we factor everything out to make an algebraic expression of parabolas reflective props and proving this. I know when the distance from the focus point to a point (x,y)on the parabola y=x^2/4*p is equal to the distance from the focus point to the slope of the tangent's y-intercept. Then the ray of light will have the same angle as the angle of the slope of the tangent and the ray of light will bounce off parallel to the axis. So when we use the pythagoreon theorem to find the length of the right triangle formed by the distance from the focus point and point (x,y) of the parabola I cant figure out what happens when we substitute the term x^2 for our 4*p*y. What our we doing algebraically. I know where replacing all of the x values for our y values. But are we giving all of the x values of the parabola a relationship to the y values that have been multiplied by 4 and p (the y value of the focus) and by the y values of the parabola point? Or does this just show/prove that the distance from the focus point to the point(x,y) of the parabola will equal the y intercept of the slope of the tangent's distance.? I hope I made it clear enough, ive been giving this a lot of thought,sorry if some of you get confused. And I apologize for the horrible spelling.
 December 11th, 2013, 03:09 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Calculus of Parabola Term Question I don't know how much lenience you have in working this problem, but if it were me, I would take a cross-section of the paraboloid with a plane containing the x and z axes, so that y = 0, and we have z = cx², thereby reducing the problem to an equivalent 2-dimensional version. I will assume the incoming light rays are all parallel to the axis of symmetry of the parabola, or traveling vertically in the negative direction. Now pick an arbitrary point $$$x_0,z_0$$=$$x_0,cx_0^2$$$ on the parabola where a ray of light is reflected. We may use a point in the first quadrant, since by symmetry, the behavior will be mirrored in the second quadrant. The law of reflection states that the angle of incidence is equal to the angle of reflection. Let $z_r= mx + b$ represent the path of the reflected ray. We need to show that b, the focus, is independent of $$$x_0,z_0$$$. The law of reflection gives us: $\frac{\pi}{2}-\tan^{\small{-1}}$$\frac{dz}{dx}$$=\tan^{\small{-1}}$$\frac{dz}{dx}$$+\pi-\tan^{\small{-1}}$$m$$$ Simplifying and using $\frac{dz}{dx}=2cx_0$, we have: $\tan^{\small{-1}}$$m$$-\frac{\pi}{2}=2\tan^{\small{-1}}$$2cx_0$$$ Taking the tangent of both sides, using a co-function identity on the left and a double-angle identity on the right, we have: $-\frac{1}{m}=\frac{4cx_0}{1-$$2cx_0$$^2}$ $m=\frac{$$2cx_0$$^2-1}{4cx_0}$ Knowing the slope m and a point on the line $z_r$, we find the z-intercept as: $b=z_0-mx_0=z_0-$$\frac{\(2cx_0$$^2-1}{4cx_0}\)x_0=\frac{4cz_0-$$2cx_0$$^2+1}{4c}$ Recall, we may write $z_0=cx_0^2$ giving us: $b=\frac{4c^2x_0^2-4c^2x_0^2+1}{4c}=\frac{1}{4c}$ Thus, we have shown that all incoming rays parallel to the z-axis and traveling towards the xy plane will be reflected through the point $Q$$0,0,\frac{1}{4c}$$$.

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