Finding parametric equations of the tangent line to a curve of intersection

Question

The question asks to find the parametric equations of the tangent line to the curve of intersection of the surface $z=2\sqrt{9-\frac{x^2}{2}-y^2}$ and the plane $x=2$ at the point $(2,\sqrt{3},4)$

I've seen solutions to questions similar to this one but they use the gradient notation (the triangle) which I am unfamiliar with. How do you do this question? I was thinking of starting with the partial derivatives w.r.t $x$ and $y$ but I'm not sure where to go from there.

Then I thought of ignoring the partial derivative of $x$ since the intersection involves $x=2$ so $x$ is unchanging, so there's that.

Thanks!

Answer 1

The parametric equation of the curve is $$ \vec v=(x,y,z)=(2,t,2\sqrt{7-t^2}) $$ so, deriving with respect to $t$, we find

$$ \frac {d\vec v}{dt}=\left(0,1,\frac{-2t}{\sqrt{7-t^2}}\right) $$ and the tangent vector at the point $(2,\sqrt{3},4)$ is $\vec u=(0,1,-\sqrt{3})$.

From this you can find the equation of the line with this direction and passing thorough the given point.

Answer 2

The gradient will give a vector normal to the surface (when you treat it as a level surface.) But you can get normal vectors in other ways. A vector normal to the plane $x=2$ is $\mathbf{i}$. A vector normal to the surface $z=f(x,y)$ is given by $-f_x\mathbf{i}-f_y\mathbf{j}+\mathbf{k}$ evaluated at your point.

A vector parallel to the tangent line is the cross product of these vectors.