Key Concepts and Formulas

• Implicit Differentiation: Used to find $\frac{dy}{dx}$ when $y$ is not explicitly a function of $x$.
• Tangent and Normal Slopes: The slope of the normal is the negative reciprocal of the slope of the tangent: $m_N = -\frac{1}{m_T}$.
• Point-Slope Form of a Line: $y - y_1 = m(x - x_1)$.
• A point on the curve satisfies the equation of the curve.

Step-by-Step Solution

Step 1: Find $\frac{dy}{dx}$ using implicit differentiation.

We are given the equation $y^2 - 3x^2 + y + 10 = 0$. We need to differentiate this with respect to $x$ to find the slope of the tangent.

$\frac{d}{dx}(y^2 - 3x^2 + y + 10) = \frac{d}{dx}(0)$ $2y\frac{dy}{dx} - 6x + \frac{dy}{dx} + 0 = 0$ $(2y + 1)\frac{dy}{dx} = 6x$ $\frac{dy}{dx} = \frac{6x}{2y + 1}$ This expression gives the slope of the tangent at any point $(x, y)$ on the curve.

Step 2: Find the slope of the normal at a point P$(x_1, y_1)$.

Let P$(x_1, y_1)$ be a point on the curve. The slope of the tangent at P is: $m_T = \frac{6x_1}{2y_1 + 1}$ The slope of the normal at P is the negative reciprocal of the tangent's slope: $m_N = -\frac{1}{m_T} = -\frac{2y_1 + 1}{6x_1}$ We assume $x_1 \neq 0$ and $2y_1 + 1 \neq 0$ for the normal slope to be defined.

Step 3: Write the equation of the normal line.

Using the point-slope form of a line, the equation of the normal at P$(x_1, y_1)$ is: $y - y_1 = m_N(x - x_1)$ Substituting the expression for $m_N$: $y - y_1 = -\frac{2y_1 + 1}{6x_1}(x - x_1)$

Step 4: Use the given y-intercept to find $y_1$.

The normal intersects the y-axis at $(0, \frac{3}{2})$. Substituting $x = 0$ and $y = \frac{3}{2}$ into the equation of the normal: $\frac{3}{2} - y_1 = -\frac{2y_1 + 1}{6x_1}(0 - x_1)$ $\frac{3}{2} - y_1 = \frac{2y_1 + 1}{6}$ Multiplying both sides by 6: $9 - 6y_1 = 2y_1 + 1$ $8 = 8y_1$ $y_1 = 1$

Step 5: Find $x_1$ using the curve's equation.

Since P$(x_1, y_1)$ lies on the curve $y^2 - 3x^2 + y + 10 = 0$, we substitute $y_1 = 1$: $(1)^2 - 3x_1^2 + (1) + 10 = 0$ $1 - 3x_1^2 + 1 + 10 = 0$ $12 - 3x_1^2 = 0$ $3x_1^2 = 12$ $x_1^2 = 4$ $x_1 = \pm 2$ So, the point P is either $(2, 1)$ or $(-2, 1)$.

Step 6: Calculate $|m|$, the magnitude of the tangent's slope.

The slope of the tangent at P$(x_1, y_1)$ is $m = \frac{6x_1}{2y_1 + 1}$. Since $y_1 = 1$, we have: $m = \frac{6x_1}{2(1) + 1} = \frac{6x_1}{3} = 2x_1$ If $x_1 = 2$, then $m = 2(2) = 4$. If $x_1 = -2$, then $m = 2(-2) = -4$. In either case, $|m| = |4| = |-4| = 4$.

Common Mistakes & Tips

• Remember the chain rule when differentiating implicitly.
• Don't forget both positive and negative roots when solving $x_1^2 = 4$.
• The slope of the normal is the negative reciprocal of the tangent's slope.

Summary

We used implicit differentiation to find the derivative of the curve, which represents the slope of the tangent. We then found the slope of the normal and used the given y-intercept of the normal to solve for the coordinates of the point P. Finally, we calculated the magnitude of the slope of the tangent at P.

The final answer is \boxed{4}.