Find the equation of the plane through a point which is perpendicular to a curve

Question

Find the equation of the plane through the point $(1, -1, 2)$ which is perpendicular to the curve of intersection of the two surfaces $x^2 + y^2 - z = 0$ and $2x^2 + 3y^2 + z^2 - 9 = 0$.

And would you be able to explain the steps? thanks!

Answer 1

First we need to find the tangent vector to the curve. We can solve the first equation for $z$ to get $$z=x^2+y^2$$ Plug this into the second equation to get that the curve satisfies $$2x^2+3y^2+x^4+2x^2y^2+y^4-9=0$$ Use implicit differentiation to obtain an equation for $\frac{dy}{dx}$ and plug in the coordinates for the point. Solve to get that $$\frac{dy}{dx}=2$$ Hence $$\frac{dz}{dx}=2x+2y \frac{dy}{dx}=2-4= -2$$ And a tangent vector is $$(1,2,-2)$$ The plane is orthogonal to this and passes through $(1,-1,2)$, hence it is given by $$(x-1)+2(y+1)-2(z-2)=0$$