Let the normal at a point P on the curve $y^2-3x^2+y+10=0$ intersect the y-axis at $\left(0,\frac{3}{2}\right)$. If m is slope of the tangent at P to the curve, then $\left|m\right|$ is equal to ________

Let $P\left(x_1,y_1\right)$ be a point on the curve $y^2-3x^2+y+10=0.$

$\frac{dy}{dx}=\frac{6x}{2y+1}$Slope of normal at $P=\frac{-\left(2y_1+1\right)}{6x_1}$

Normal at P meets y-axis at $Q\left(0,\frac{3}{2}\right)$

Slope of normal at P= slope of PQ

$\begin{array}{l}\frac{-\left(2y_1+1\right)}{6x_1}=\frac{y_1-\frac{3}{2}}{x_1-0}\\ -2y_1-1=6y_1-9\\ 8y_1=8\\ y_1=1\end{array}$
$$\begin{gathered} y_1=1\Rightarrow 1-3{x_1}^2+1+10=0\Rightarrow x_1=\pm 2 \\ \Rightarrow slope of the \tan gent=m=\frac{6x_1}{2y_1+1} \\ \Rightarrow \left|m\right|=\left|\frac{6\times 2}{2\times 1+1}\right|=4 \end{gathered}$$