Let the normal at a point P on the curve
Question:

Let the normal at a point $P$ on the curve $y^{2}-3 x^{2}+y+10=0$ 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 $|m|$ is equal to__________.

Solution:

$P \equiv\left(x_{1}, y_{1}\right)$

$2 y y^{\prime}-6 x+y^{\prime}=0$

$\Rightarrow \quad y^{\prime}=\left(\frac{6 x_{1}}{1+2 y_{1}}\right)$

$\left(\frac{\frac{3}{2}-y_{1}}{-x_{1}}\right)=-\left(\frac{1+2 y_{1}}{6 x_{1}}\right)$

[By point slope form, $\left.y-y_{1}=m\left(x-x_{1}\right)\right]$

$\Rightarrow 9-6 y_{1}=1+2 y_{1}$

$\Rightarrow y_{1}=1$

$\therefore \quad x_{1}=\pm 2$

$\therefore \quad$ Slope of tangent $(m)=\left(\frac{\pm 12}{3}\right)=\pm 4$

$\therefore \quad|m|=4$