Although this is a rather simple idea, I still think its pretty cool. All it is finding the locus of the point equidistant from the focus F(h, k) and the directrix x - y = 0. So far, using the directrix mx - y + b = 0, I haven't been able to achieve a parabola. I assume my equations are mistaken, so I will continue to work on those. For now I present my first diagonal parabola!
|x – y| = √((x – h)2 + (y – k)2)
√(1 + 1)
(x – y)2 = 2((x – h)2 + (y – k)2)
x2 – 2xy + y2 = 2(x2 – 2hx + h2 + y2 – 2ky + k2)
-x2 – 2xy – y2 = -4hx + 2h2 + - 4ky + 2k2
x2 + 2xy + y2 – 4hx - 4ky + 2(k2 + h2) = 0
(x – y)2 = 2((x – h)2 + (y – k)2)
x2 – 2xy + y2 = 2(x2 – 2hx + h2 + y2 – 2ky + k2)
-x2 – 2xy – y2 = -4hx + 2h2 + - 4ky + 2k2
x2 + 2xy + y2 – 4hx - 4ky + 2(k2 + h2) = 0
This gives the following parabola:
Fig. 5 The Diagonal Parabola
I've been trying to figure out an equation for a diagonal palabora the whole day but only got to x^2-2xy+y^2 haha. One thing i wanted to point out that you might have already noticed too is that this type of "diagonal palabra" equation always has the opposite palabora too, its just you have to zoom out a lot to see it in the 4th quadrant. What I found really cool was that no matter what diagonal palabra equation you will find on the internet, they always seem to have this feature. Does that mean true diagonal palabras dont exist, or have we just not found the equation. Anyways, dont know how you got this but good job.
ReplyDelete