Let $L_1, L_2$ be the lines passing through the point $P(0,1)$ and touching the parabola $9 x^2+12 x+18 y-14=0$. Let $Q$ and $R$ be the points on the lines $L_1$ and $L_2$ such that the $\triangle P Q R$ is an isosceles triangle with base $Q R$. If the slopes of the lines $Q R$ are $m_1$ and $m_2$, then $16\left(m_1^2+m_2^2\right)$ is equal to _______

Solution

$\begin{aligned} & 9 x^2+12 x+4=-18(y-1) \\ & (3 x+2)^2=-18(y-1) \\ & \left(x+\frac{2}{3}\right)^2=-2(y-1)\end{aligned}$
$\begin{aligned} & (0,1) \\ & y=m x+1 \\ & \left(x+\frac{2}{3}\right)^2=-2(y-1) \\ & (3 x+2)^2=-18 m x \\ & 9 x^2+(12+18 m) x+4=0 \\ & 4(6+9 m)^2=4(36) \\ & 6+9 m=6,-6 \\ & m=0, \frac{-4}{3}\end{aligned}$
$\begin{aligned} & \tan \theta=-\frac{4}{3} \\ & \frac{2 \tan \frac{\theta}{2}}{1-\tan ^2 \frac{\theta}{2}}=\frac{-4}{3} \\ & \left(\tan \frac{\theta}{2}-2\right)\left(2 \tan \frac{\theta}{2}+1\right)=0 \\ & \tan \frac{\theta}{2}=2, \frac{-1}{2} \\ & \mathrm{~m}_{\mathrm{QR}}=\tan \left(90+\frac{\theta}{2}\right)\end{aligned}$ $\begin{aligned} & =-\cot \frac{\theta}{2} \\ & \mathrm{~m}_1=\frac{-1}{2} \quad \mathrm{~m}_2=\frac{-1}{-1 / 2}=2 \\ & 16\left(\mathrm{~m}_1^2+\mathrm{m}_2^2\right)=16\left(\frac{1}{4}+4\right) \\ & =4+64=68\end{aligned}$

Hence, the correct answer is 68.