Question

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 the slope of the tangent at $P$ to the curve, then $\left|m\right|$ is equal to ?

Answer 1

Finding the value of $\left|m\right|$:

Given curve is $y^2-3x^2+y+10=0$.

Let $P$ be a point $\left(x_1,y_1\right)$.

Step 1: Find the slop of the Equation

Now differentiate the given curve with respect to $\text{'}x\text{'}$.

$$\begin{gathered} \Rightarrow 2yy\text{'}-6x+y\text{'}=0 \\ \Rightarrow 2yy\text{'}+y\text{'}=6x \\ \Rightarrow \left(2y+1\right)y\text{'}=6x \\ \Rightarrow y\text{'}=\frac{6x}{2y+1} \end{gathered}$$

Therefore, the slope of the tangent to the curve is $\frac{6x_1}{2y_1+1}$.

Step 2 : Determine the slope of the normal at $\left(0,\frac{3}{2}\right)$.

$$\begin{gathered} \Rightarrow y\text{'}=\frac{y_1-y_2}{x_1-x_2} \\ \Rightarrow y\text{'}=\frac{y_1-{\displaystyle \frac{3}{2}}}{x_1-0} \\ \Rightarrow y\text{'}=\frac{2y_1-3}{2x_1} \end{gathered}$$

Therefore, the slope of the normal to the curve is $\frac{2y_1-3}{2x_1}$.

Step 3: Find the value of $y_1$

$m_{\text{normal}}\times m_{\text{tangent}}=-1$

$$\begin{gathered} \Rightarrow \frac{2y_1-3}{2x_1}\times \frac{6x_1}{2y_1+1}=\begin{array}{l}-\end{array}\begin{array}{l}1\end{array} \\ \Rightarrow \frac{3\begin{array}{rcl}& & \left(2y_1-3\right)\end{array}}{\begin{array}{rcl}& & 2y_1+1\end{array}}=-1 \\ \Rightarrow 6y_1-9=-\left(2y_1+1\right) \\ \Rightarrow 8y_1=8 \\ \Rightarrow y_1=1 \end{gathered}$$

Step 4: Find the value of $x_1$

Substitute $y_1$ as $1$ in the equation ${y_1}^2-3{x_1}^2+y_1+10=0$

$$\begin{gathered} \Rightarrow 1^2-3{x_1}^2+1+10=10 \\ \Rightarrow 12-3{x_1}^2=0 \\ {\Rightarrow x_1}^2=\frac{12}{3} \\ {\Rightarrow x_1}^2=4 \\ \Rightarrow x_1=\pm 2 \end{gathered}$$

Step 5: Find the value of $\left|m\right|$

Substitute $x_1$ as $\pm 2$ and $y_1$ as $1$ in $\frac{6x_1}{2y_1+1}$.

$$\begin{gathered} m=\frac{6\left(\pm 2\right)}{2\left(1\right)+1} \\ m=\frac{\pm 12}{3} \\ m=\pm 4 \end{gathered}$$

Then $\left|m\right|=4$.

Therefore, $\left|m\right|$ is equal to $4$.