Question

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

Answer 1

Solution:

Suppose that the plane

$$2x-3y+6z=6$$

touches the conicoid

$$4x^2-9y^2+36z^2=36$$

at some point

$$M(x_0,y_0,z_0).$$

Then the plane equation can be written as follows:

$${F_x}^,|_M(x-x_0)+{F_y}^,|_M(y-y_0)+{F_z}^,|_M(z-z_0)=0,$$

where

$$F(x,y,z)=4x^2-9y^2+36z^2-36,$$

$${F_x}^,|_M=8x_0,{F_y}^,|_M=-18y_0,{F_z}^,|_M=72z_0;$$

or

$$8x_0(x-x_0)-18y_0(y-y_0)+72z_0(z-z_0)=0,$$

$$8x_0x-18y_0y+72z_0z-(8{x_0}^2-18{y_0}^2+{72z_0}^2)=0.$$

The coefficients of the last equation must be equal to the coefficients of the equation

$$2x-3y+6z-6=0.$$

Check it out:

$$8x_0=2;18y_0=3;72z_0=6;8{x_0}^2-18{y_0}^2+{72z_0}^2=6;$$

or

$$x_0=\frac{1}{4};y_0=\frac{1}{6};z_0=\frac{1}{12};\frac{1}{2}=6.$$

The last equality is wrong,so the plane

$$2x-3y+6z=6$$

does not touch the conicoid

$$4x^2-9y^2+36z^2=36$$

and there is no point of contact.

Answer:so the plane

$$2x-3y+6z=6$$

does not touch the conicoid

$$4x^2-9y^2+36z^2=36$$

and there is no point of contact.