What I did was:
Take the partial derivative of x and y obtaining [$2xy^2,2x^2y$] as the general gradient of my tangent line.
I plugged in $(1,2)$ to obtain the specific gradient of my tangent line, obtaining $[8,4]$.
- We know that if the dot product of two vectors = 0, then these two vectors are perpendicular. So I solved for the perpendicular vectors and obtained: $[-4,8]$ or $[4,-8]$ as two specific vectors that are perpendicular to the implicit curve at $(1,2)$.
I am going to feel like a total idiot with tunnel vision but up until literally writing this post I have been trying to solve for the perpendicular vector to the line tangent to my implicit curve at the point $(1,2)$.
Would the vector perpendicular to the implicit curve of $(xy)^2 - 4 = 0$ at the point $(1,2)$ just be the specific gradient of my tangent line at the point $(1,2)$. So the answer to my question should just be $[8,4]$?
I appreciate the help.
The equation is $(xy)^2=4$, thus $xy=\pm2$, the union of two hyperbolas. Your point lies on the branch $xy=2$, so you’re asking for the normal direction to the graph of the function $y=2/x$ where $x=1$. Differentiating, we have $\frac{dy}{dx}=-2/x^2$, so the tangent at your point has slope $-2$, and the normal line has slope $1/2$, so a good normal vector would be $[2,1]$, agreeing with your final result.