If the tangent at $P\left(t_1\right)$ on the curve $y=8t^3-1$ ,  $x=4t^2+1$ is normal at $Q\left(t_2\right)$   then $\frac{t_1}{t_2}=$

$$\begin{gathered} P(4t_1^2+1,8t_1^3-1) Q(4t_2^2+1,8t_2^3-1) \\ \frac{dy}{dx}=\frac{24t^2}{8t}=3t \end{gathered}$$

Slope of tangent at  $P$ is  $m_1=3t_1$

Slope of tangent at $Q$ is  $m_2=3t_2$

$$\begin{gathered} m_1{m}_2=-1 \\ 9t_1{t}_2=-1 ----\left(1\right) \end{gathered}$$

Eq. of tangent at  $P$ is  $3t_{1}x-y=4t_1^3+3t_1+1$
If passes through  $Q$
$\therefore {\left(\frac{t_1}{t_2}\right)}^3-3\left(\frac{t_1}{t_2}\right)+2=0$

$\therefore \frac{t_1}{t_2}=-2 ----\left(2\right)$

From  $\left(1\right) \& \left(2\right) t_1=-\frac{\sqrt{2}}{3}$

The normal at  $Q$ is tangent at the point  $P$

$\therefore 27(\sqrt{2}x+y)=35\sqrt{2}-27$