Answer to Question #91516 in Analytic Geometry for Ra

Question

Prove that the plane 2x-3y+6z=6 touches the conicoid 4x²-9y²+36z²=36
Find the point of contact.

Answer 1

1. Represent the plain and the conicoid in an implicit form:

$F(x,y,z)=2x-3y+6z-6=0;$

$G(x,y,z)=4x²-9y²+36z²-36=0.$

2. The normal vectors to the surfaces are determined by the partial derivatives:

$N_F = (\frac{∂F}{∂x},\frac{∂F}{∂y},\frac{∂F}{∂z}) =(2,-3,6);$  

$N_G = (\frac{∂G}{∂x},\frac{∂G}{∂y},\frac{∂G}{∂z}) =(8x,-18y,72z).$

3. If the plain touches the conicoid, then their normal vectors at the point of contact must be collinear:

$N_G = k * N_F;$

$(8x,-18y,72z)=k*(2,-3,6);$

$x=\frac{k}{4}; y=\frac{k}{6}; z=\frac{k}{12}.$

4. Put the values of variables to the equation of the plain and find the coordinates of the contact point:

$2*\frac{k}{4}-3*\frac{k}{6}+6*\frac{k}{12}-6=0;$

$\frac{k}{2}-\frac{k}{2}+\frac{k}{2}-6=0;$

$k=12;x=3;y=2;z=1;$ M₀(3,2,1).

This point lies also in the conicoid:

$4x²-9y²+36z²-36=4*3²-9*2²+36*1²-36=0.$

5. Thus, M₀ is a common point of surfaces, and normal vectors are collinear there, so the plain touches the conicoid at the point M₀(3,2,1).