I've been provided with this problem to solve:
Find the equation of the line tangent to the surface $ z = f(x, y) = x^2 - y^2 $ at the point $ (2, 3) $ and in the direction of the unit vector $ {\mathbf u} = {\langle \frac{\sqrt 2}{2}}, \frac{\sqrt 2}{2} \rangle $.
So far, I've already determined the plane tangent to $ (2, 3) $ using the equation $ f_x(2, 3) + f_y(2, 3) = z - f(2, 3) $.
$$ f(2, 3) = 2^2 - 3^2 = -5 $$
$$ f_x(x, y) = \frac{\partial}{\partial x}(x^2 - y^2) = 2x \to f_x(2, 3) = 2 \times 2 = 4 $$
$$ f_y(x, y) = \frac{\partial}{\partial y}(x^2 - y^2) = -2y \to f_y(2, 3) = -2 \times 3 = -6 $$
The plane is
$$ 4(x - 2) - 6(y - 3) - (z + 5) = 0 $$
I also determined the directional derivative using the equation $ D_{\mathbf u}f(2, 3) = \nabla f(2, 3) \cdot {\mathbf u} = \langle f_x(2, 3), f_y(2, 3) \rangle \cdot {\mathbf u} $.
$$ D_{\mathbf u}f(2, 3) = \langle 4, -6 \rangle \cdot \langle \frac{\sqrt 2}{2}, \frac{\sqrt 2}{2} \rangle = -\sqrt 2 $$
From the equation of the plane tangent to $ (2, 3) $, the parametric equations for the normal line to the plane are
$$ x(t) = 2 + 4t $$ $$ y(t) = 3 + 6t $$ $$ z(t) = -5 - t $$
Is there a way to yield the parametric equations for the particular line based on $ D_{\mathbf u}f(2, 3) $?