Question

Show that the plane 3x+12y=62-17 = 0, touches the conicoid 3x

Answer 1

Show that the plane $3x+12y-6z-17 = 0,$ touches the conicoid $3x^2-6y^2+9z^2+17=0$ and find the point of contact.

$$F(x, y, z)=3x^2-6y^2+9z^2+17=0$$

$$F_x=6x, F_y=-12y, F_z=18z$$

The equation of the tangent plane is

$$F_x(x-x_0)+F_y(y-y_0)+F_z(z-z_0)=0$$

Substitute

$$6x_0(x-x_0)-12y_0(y-y_0)+18z_0(z-z_0)=0$$

$$3x_0(x-x_0)-6y_0(y-y_0)+9z_0(z-z_0)=0$$

$$3x_0x-6y_0y+9z_0z-(3x_0^2-6y_0^2+9z_0^2)=0$$

$$3x_0x-6y_0y+9z_0z+17=0$$

$$-3x_0x+6y_0y-9z_0z-17=0$$

Given the plane $3x+12y-6z-17 = 0.$ Then

$$x_0=-1, y_0=2,z_0=2/3$$

The plane $3x+12y-6z-17 = 0,$ touches the conicoid $3x^2-6y^2+9z^2+17=0$ at the point $(-1,2,2/3).$