PROBLEM: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?

Intersecting normal:

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.