Question

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

Answer 1

Let co-ordinates of $P$ be $(x_1,y_1)$
Differentiating the curve w.r.t. $x$
$2y{y}^{\prime }-6x+y^{\prime }=0$
Slope of tangent at $P(x_1,y_1)$ is
$y^{\prime }=\frac{6x_1}{1+2y_1}$

$m_{\text{normal}}=\frac{y_1-\frac{3}{2}}{x_1-0}$
$\because m_{\text{normal}}\times m=-1$
$\Rightarrow \frac{y_1-\frac{3}{2}}{x_1}\times \frac{6x_1}{1+2y_1}=-1$
$\Rightarrow y_1=1$
$\Rightarrow x_1=\pm 2$
Therefore, slope of tangent $=\pm \frac{12}{3}=\pm 4$
$\Rightarrow |m|=4$