Question

Find an equation of the straight line tangent, at the given point, to the level curve of the given function passing through that point.
f(x, y) = x^{2}- y^{2} at (2, -1)

Answer 1

$f(2,-1)=2^2-(-1)^2=3$

so we're considering the level curve

$x^2-y^2=3$

The tangent line to this curve at (2, -1) will be the value of $\dfrac{\mathrm dy}{\mathrm dx}$ at this point. Differentiating yields

$2x-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac xy$

and so the slope of the tangent would be $\dfrac2{-1}=-2$.

The tangent line then has equation

$y-(-1)=-2(x-2)\implies y=-2x+3$