at ,
normal line thru is ...
substitute for in the original curve equation ...
PROBLEM:
Intersecting normal: The line that is normal to the curve at intersects the curve at what other point?
ATTEMPT:
My biggest problem lies in finding the intersection between a single-valued function and a multi-valued function. Finding the intersection between a single valued function and the same is easy enough, so I figured if I could express as a series of single valued functions, then I should be able to find where the normal intersects. At least in theory. With some investigation, I realized I didn't know how to properly express the curve, , as such. I thought that using the quadratic formula might be appropriate for this operation since the curve is a two-valued function specifically, which resulted in the expression: But when I tried finding the pair, it resulted in and which doesn't agree with what I expected from the graph.
I've determined that the normal curve is the function, .
I appreciate any thoughts you have on this.
Thanks for reading.
You can also do it like this
x2+2xy-3y2=0
x2+2xy+y2-y2-3y2=0
(x+y)2-4y2=0
(x+y)2=(2y)2
The normal line to x=y and passing through (1,1) has the equation y=2-x
y=2-x intersects x=-3y at (3,-1)
Thus the required point is (3,-1)
x+y=±2y
x=y, x=-3y