I've been given the following information: function: $f(x,y)=\frac{x}{\sqrt{x^2-2y^2}}$ Level curve defined by $f(x,y)=c$, where $c$ is a constant
Now I have to find the tangent lines of the level curve when $c=\sqrt{2}$ and $x=1$ Not sure how to set this up.
Suppose a tangent line through $(1,\ y_0)$ has the equation $(x(t),\ y(t)) = (at + 1,\ bt+y_0)$.
You have $0 = \frac{d}{dt}f(x(t),\ y(t))|_{(0,0)} = \frac{\partial f}{\partial x}(1,\ y_0) \frac{dx}{dt}(0) + \frac{\partial f}{\partial y}(1,\ y_0) \frac{dy}{dt}(0)$.
In other words, $0 = a\frac{\partial f}{\partial x}(1,\ y_0) + b\frac{\partial f}{\partial y}(1,\ y_0)$.
This all becomes concrete if you know the specific values of $y_0$ to plug in.