The normal to the curves $y^2-3x^2+y+10=0$ at a point $P$ passes through the point $(0, 3/2).$ lf the slope of the tangent to the curve at $P$ is $m,$ then the value of $\left|m\right|$ is

Let the coordinates of $P$ be $(x_1, y_1).$ The equation of the curve is

      $y^2-3x^2+y+10=0$

Differentiating with respect to $x,$ we obtain

    $2y\frac{dy}{dx}-6x+\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=\frac{6x}{2y+1}\Rightarrow {\left(\frac{dy}{dx}\right)}_P=\frac{6x_1}{2y_1+1}$

The equation of the normal of $P (x_1, y_1)$ is

     $y-y_1=-\frac{2y_1+1}{6x_1}\left(x-x_1\right)$

It passes through $(0, 3/2).$

$\therefore \frac{3}{2}-y_1=-\frac{2y_1+1}{6x_1}\left(0-x_1\right)$

$\Rightarrow \frac{3}{2}-y_1=\frac{2y_1}{6}+1\Rightarrow 9-6y_1=2y_1+1\Rightarrow y_1=1$

$P (x_1, y_1)$ lies on the curve $y^2-3x^2+y+10=0.$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{y}_1^2-3x_1^2+y_1+10=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}1-3x_1^2+1+10=0\Rightarrow x_1=\pm 2\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\left(\frac{dy}{dx}\right)}_p=\frac{\pm 12}{3}=\pm 4\Rightarrow m=\pm 4\end{array}$

Hence, $\left|m\right|=4.$