Question #200475

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


1
Expert's answer
2022-02-01T00:59:08-0500

Suppose that the plane



2x3y+6z=62x-3y+6z=6

touches the conicoid



4x29y2+36z2=364x^2-9y^2+36z^2=36

at some point



M(x0,y0,z0).M(x_0,y_0,z_0).

Then the plane equation can be written as follows:



Fx,M(xx0)+Fy,M(yy0)+Fz,M(zz0)=0,{F_x}^,|_M(x-x_0)+{F_y}^,|_M(y-y_0)+{F_z}^,|_M(z-z_0)=0,

where



F(x,y,z)=4x29y2+36z236,F(x,y,z)=4x^2-9y^2+36z^2-36,Fx,M=8x0,Fy,M=18y0,Fz,M=72z0;{F_x}^,|_M=8x_0,{F_y}^,|_M=-18y_0,{F_z}^,|_M=72z_0;

or



8x0(xx0)18y0(yy0)+72z0(zz0)=0,8x_0(x-x_0)-18y_0(y-y_0)+72z_0(z-z_0)=0,8x0x18y0y+72z0z(8x0218y02+72z02)=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



2x3y+6z6=0.2x-3y+6z-6=0.

Check it out:



8x0=2;18y0=3;72z0=6;8x0218y02+72z02=6;8x_0=2;18y_0=3;72z_0=6;8{x_0}^2-18{y_0}^2+{72z_0}^2=6;

or


x0=14;y0=16;z0=112;12=6.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



2x3y+6z=62x-3y+6z=6

does not touch the conicoid



4x29y2+36z2=364x^2-9y^2+36z^2=36

and there is no point of contact.


Answer: so the plane



2x3y+6z=62x-3y+6z=6


does not touch the conicoid



4x29y2+36z2=364x^2-9y^2+36z^2=36

and there is no point of contact.


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