Question

Find the equations of the tangent planes to the conicoid 7x^2-3y^2-z^2+21=0, which pass through the line 7x-6y+9=0, z=3

Answer 1

The equation of the line is $\begin{cases} 7x-6y+9=0 \\ z=3 \end{cases} \quad \begin{cases} x=\frac{6t-9}{7} \\ y=t \\ z=3 \end{cases}$

If the plane $ax+by+cz+d=0$ passes through this line, then

$a\frac{6t-9}{7}+bt+3c+d=(6a/7+b)t+(3c+d-9a/7)=0$

It means, that $\begin{cases} 6a/7+b=0 \\ 3c+d-9a/7=0 \end{cases} \quad \begin{cases} a=7\alpha \\ b=-6\alpha \\ c=\beta \\ d=9\alpha-3\beta \end{cases}$

So, we have the equation of this plane $7\alpha x-6\alpha y+\beta z+(9\alpha -3\beta)=0$

Let $7\alpha x-6\alpha y+\beta z+(9\alpha -3\beta)=0$ be the equation of the tangent plane to the conicoid $7x^2-3y^2-z^2+21=0$ at the point $P(x_0,y_0,z_0)$ .

The point $P(x_0,y_0,z_0)$ is on the conicoid: $7x_0^2-3y_0^2-z_0^2+21=0$ .

The equation of the tangent plane is $f_x|_P(x-x_0) +f_y|_P(y-y_0)+f_z|_P(z-z_0)=0$ , where $f(x,y,z)=7x^2-3y^2-z^2+21$

We have $14x_0(x-x_0)-6y_0(y-y_0)-2z_0(z-z_0)=0$

$2(7xx_0-3yy_0-zz_0+21)-2(7x_0^2-3y_0^2-z_0^2+21)=0$

Therefore, $7xx_0-3yy_0-zz_0+21=0$ is the equation of the tangent plane at $P(x_0,y_0,z_0)$ .

Using the fact that $7xx_0-3yy_0-zz_0+21=0$ and $7\alpha x-6\alpha y+\beta z+(9\alpha -3\beta)=0$ are equations of the same plane, we get

$\begin{cases} x_0=\alpha \\ y_0=2\alpha \\ z_0=-\beta \\ 9\alpha -3\beta=21 \end{cases} \quad \begin{cases} x_0=\alpha \\ y_0=2\alpha \\ z_0=7-3\alpha \\ \beta=3\alpha -7 \end{cases}$

Substituting $(x_0,y_0,z_0)$ into $f(x_0,y_0,z_0)=0$ , we get $7\alpha ^2-12\alpha ^2-(3\alpha-7)^2+21=-14\alpha^2+42\alpha -28=0$

There are 2 roots of this equation: $\alpha _1=1$ , $\alpha_2 =2$ . Then we get $\beta_1=-4$ and $\beta_2=-1$ .

Therefore, we have two planes: $7x-6y-4z+21=0$ and $14x-12y-z+21=0$ .

Answer: $7x-6y-4z+21=0$ and $14x-12y-z+21=0.$