# Thread: Finding the intersection between the normal line and an implicit irrational function

1. ## Finding the intersection between the normal line and an implicit irrational function

PROBLEM:

Intersecting normal:
The line that is normal to the curve $x^2+2xy-3y^2=0$ at $(1,1)$ 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
$x^2+2xy-3y^2=0$ 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, $x^2+2xy-3y^2=0$, 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: $y=\frac{x}{3}\pm\sqrt{16}x.$ But when I tried finding the pair, it resulted in $(\frac{7}{53},\frac{343}{159})$ and $(\frac{7}{43},\frac{329}{129})$ which doesn't agree with what I expected from the graph.

I've determined that the normal curve is the function, $y=-\frac{4}{3}x+\frac{7}{3}$.

I appreciate any thoughts you have on this.

2. ## Re: Finding the intersection between the normal line and an implicit irrational funct

$\frac{d}{dx}(x^2 + 2xy - 3y^2 = 0)$

$\frac{dy}{dx} = \frac{x+y}{3y-x}$

at $(1,1)$ , $\frac{dy}{dx} = 1$

normal line thru $(1,1)$ is ...

$y - 1 = -1(x-1)$

$y = 2-x$

substitute $(2-x)$ for $y$ in the original curve equation ...

3. ## Re: Finding the intersection between the normal line and an implicit irrational funct

Algebra error:
$y={2x\pm \sqrt{4x^2+12x^2}\over 6}={x\over 3}\pm {2x\over 3}$
As pointed out above, the normal line has equation $y=2-x$. Easy algebra shows the other point of intersection is (3,-1).

4. ## Re: Finding the intersection between the normal line and an implicit irrational funct

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