Finding the intersection between the normal line and an implicit irrational function

**PROBLEM:**

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

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

I appreciate any thoughts you have on this.

Thanks for reading.

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

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

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

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

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

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

$\displaystyle y = 2-x$

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

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

Algebra error:

$\displaystyle y={2x\pm \sqrt{4x^2+12x^2}\over 6}={x\over 3}\pm {2x\over 3}$

As pointed out above, the normal line has equation $\displaystyle y=2-x$. Easy algebra shows the other point of intersection is (3,-1).

Attachment 26494

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