The plane 3x+4y+2z=1 touches the conicoid 3x2+2y2+z2=1.
True or false
Solution:
Suppose that the plane
3x+4y+2z=1
touches the conicoid
3x2+2y2+z2=1
at some point
M(x0,y0,z0).
Then the plane equation can be written as follows:
Fx′∣M(x−x0)+Fy′∣M(y−y0)+Fz′∣M(z−z0)=0.
where
F(x,y,z)=3x2+2y2+z2−1,
Fx′∣M=6x0, Fy′∣M=4y0, Fz′∣M=2z0
or
6x0(x−x0)+4y0(y−y0)+2z0(z−z0=0),
6x0x+4y0y+2z0z−(6x02+4y02+2z02)=0.
The coefficients of the last equation must be equal to the coefficients of the equation
3x+4y+2z−1=0.
Check it out:
6x0=3, 4y0=4, 2z0=2, 6x02+4y02+2z02=1.
x0=21, y0=1, z0=1, 7.5=1.
The last equality is wrong,
Answer:
so the plane
3x+4y+2z=1
does not touch the conicoid
3x2+2y2+z2=1
and there is no point of contact.
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