If  a tangent drawn to the curve $f\left(x\right)=x^3-9x-1$  at $P(x_0,f\left(x_0\right))$  meets the curve again at Q,  $m_A$ denotes  the slope of the tangent at A and  $m_{OB}$ denotes the slope of  line joining origin O and a point B on the curve then

$$\begin{gathered} Given, f\left(x\right)=x^3-9x-1 \\ {f}^{\text{'}}\left(x\right)=3x^2-9 \\ m=f^1\left(x_0\right)=3x_0^2-9 \\ Let, Q is (x_1,y_1);y_1=x_1^3-9x_1-1 \\ Also, \\ 3x_0^2-9=\frac{y_1-f\left(x_0\right)}{x_1-x_0} \\ 3x_0^2-9=\frac{(x_1^3-9x_1-1)-(x_0^3-9x_0-1)}{(x_1-x_0)} \\ =\frac{(x_1^3-x_0^3)-9(x_1-x_0)}{(x_1-x_0)} \\ =\frac{{(x_1-x_0)}^3+3x_1{x}_0(x_1-x_0)-9(x_1-x_0)}{(x_1-x_0)} \\ ={(x_1-x_0)}^2+3x_1{x}_0-9 \\ 3x_0^2=x_1^2+x_1{x}_0+x_0^2 \\ {x}_1^2+x_1{x}_0-2x_0^2=0 \\ {x}_1^2+2x_1{x}_0-x_1{x}_0-2x_0^2=0 \\ {x}_1(x_1+2x_0)-x_0(x_1+2x_0)=0 \\ (x_1-x_0)(x_1+2x_0)=0 \\ {x}_1=x_{0}or{x}_1=-2x_0 \end{gathered}$$

Thus,  $x_1=-2x_0$

$$\begin{gathered} m_Q=3x_1^2-9=3{(-2x_0)}^2-9=12x_0^2-9 \\ {m}_p=3x_0^2-9 \\ {m}_Q-4m_p=12x_0^2-9-4(3x_0^2-9)=27 \end{gathered}$$

$m_{op}=\frac{f\left(x_0\right)-0}{x_0-0}=\frac{(x_0^3-9x_0-1)}{x_0}=-9at{x}_0=1$

$$\begin{gathered} m_{OQ}=\frac{y_1-0}{x_1-0}=\frac{(x_1^3-9x_1-1)}{x_1} \\ =\frac{(-8+18-1)}{(-2)}at{x}_1=-2x_0=-2 \\ =-\frac{9}{2} \end{gathered}$$