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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)

Verified Answer :

LammettHash

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

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

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

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