Question

The plane 3x+4y+2z=1 touches the conicoid 3x^2+2y^2+z^2=1
True or false with full explanation

Answer 1

The plane $3x+4y+2z=1$ touches the conicoid $3x^2+2y^2+z^2=1.$

True or false

Solution:

Suppose that the plane

$3x+4y+2z=1$

touches the conicoid

$3x^2+2y^2+z^2=1$

at some point

$M(x_0,y_0,z_0).$

Then the plane equation can be written as follows:

$F_{x'}\mid_M(x-x_0)+F_{y'}\mid_M(y-y_0)+F_{z'}\mid_M(z-z_0)=0.$

where

$F(x,y,z)=3x^2+2y^2+z^2-1,$

$F_{x'}\mid_M=6x_0,\space F_{y'}|_M=4y_0,\space F_{z'}|_M=2z_0$

or

$6x_0(x-x_0)+4y_0(y-y_0)+2z_0(z-z_0=0),$

$6x_0x+4y_0y+2z_0z-(6x_0^2+4y_0^2+2z_0^2)=0.$

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

$3x+4y+2z-1=0.$

Check it out:

$6x_0=3,\space 4y_0=4,\space 2z_0=2,\space 6x_0^2+4y_0^2+2z_0^2=1.$

$x_0=\frac{1}{2},\space y_0=1,\space z_0=1,\space 7.5=1.$

The last equality is wrong,

Answer:

so the plane

$3x+4y+2z=1$

does not touch the conicoid

$3x^2+2y^2+z^2=1$

and there is no point of contact.