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
`(\mathrm dy)/(\mathrm dx)` at this point. Differentiating yields


`2x-2y(\mathrm dy)/(\mathrm dx)=0\implies(\mathrm dy)/(\mathrm dx)=\frac xy`

and so the slope of the tangent would be
`\frac2{-1}=-2`.

The tangent line then has equation


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