Suppose that the plane
2x−3y+6z=6touches the conicoid
4x2−9y2+36z2=36at 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)=4x2−9y2+36z2−36,Fx,∣M=8x0,Fy,∣M=−18y0,Fz,∣M=72z0;or
8x0(x−x0)−18y0(y−y0)+72z0(z−z0)=0,8x0x−18y0y+72z0z−(8x02−18y02+72z02)=0.The coefficients of the last equation must be equal to the coefficients of the equation
2x−3y+6z−6=0.Check it out:
8x0=2;18y0=3;72z0=6;8x02−18y02+72z02=6;or
x0=41;y0=61;z0=121;21=6.The last equality is wrong,so the plane
2x−3y+6z=6does not touch the conicoid
4x2−9y2+36z2=36and there is no point of contact.
Answer: so the plane
2x−3y+6z=6
does not touch the conicoid
4x2−9y2+36z2=36and there is no point of contact.
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