Let the normal at a point p on the curve ${\mathrm{y}}^2-3{\mathrm{x}}^2+\mathrm{y}+10=0$ intersect the y-axis at $\left(0,\frac{3}{2}\right)$ if ‘m’ is the slope of the tangent at P to the curve, then $\left|\mathrm{m}\right|=\_\_\_\_\_\_$

${\mathrm{y}}^2-3{\mathrm{x}}^2+\mathrm{y}+10=0$ intersect y-axis at $\left(0,\frac{3}{2}\right)$
Diff. w.r.t. ‘x’
$\begin{array}{l}2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}-6\mathrm{x}+\frac{\mathrm{dy}}{\mathrm{dx}}=0\\ \frac{\mathrm{dy}}{\mathrm{dx}}(2\mathrm{y}+1)=6\mathrm{x}\\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{6\mathrm{x}}{2\mathrm{y}+1}\end{array}$
Equation of normal is  $\mathrm{y}-\frac{3}{2}=\frac{-\left(2\mathrm{y}+1\right)}{6\mathrm{x}}\left(\mathrm{x}-0\right)$
$\begin{array}{l}9-6{\mathrm{y}}_1=1+2{\mathrm{y}}_1\\ \Rightarrow {\mathrm{y}}_1=1\\ \Rightarrow {\mathrm{x}}_1=\pm 2\end{array}$
$\therefore$slope of tangent$\left(\frac{+12}{3}\right)=\pm 4$
$\left|\mathrm{m}\right|=4$