We are tasked with finding the equations of the tangent and normal lines to the given curve $x^3 + y^2 + 2x - 6 = 0$ at the point $M(-1, 3)$.

Steps:

1. Verify that $M(-1, 3)$ lies on the curve: Substitute $x = -1$ and $y = 3$ into the curve equation: $(-1)^3 + (3)^2 + 2(-1) - 6 = -1 + 9 - 2 - 6 = 0$ Therefore, $M(-1, 3)$ is indeed a point on the curve.

2. Differentiate the curve equation to find the slope of the tangent: The given curve is $x^3 + y^2 + 2x - 6 = 0$. Differentiating both sides with respect to $x$, we get: $3x^2 + 2y \frac{dy}{dx} + 2 = 0$ Solving for $\frac{dy}{dx}$ (the slope of the tangent): $\frac{dy}{dx} = -\frac{3x^2 + 2}{2y}$

3. Find the slope of the tangent at $M(-1, 3)$: Substitute $x = -1$ and $y = 3$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{3(-1)^2 + 2}{2(3)} = -\frac{3 + 2}{6} = -\frac{5}{6}$ So, the slope of the tangent is $m_{\text{tangent}} = -\frac{5}{6}$.

4. Write the equation of the tangent line: The equation of a line is given by: $y - y_1 = m(x - x_1)$ Substituting $m = -\frac{5}{6}$, $x_1 = -1$, and $y_1 = 3$: $y - 3 = -\frac{5}{6}(x + 1)$ Simplify: $y = -\frac{5}{6}x - \frac{5}{6} + 3$ $y = -\frac{5}{6}x + \frac{13}{6}$ So, the equation of the tangent line is: $y = -\frac{5}{6}x + \frac{13}{6}$

5. Find the slope of the normal line: The slope of the normal line is the negative reciprocal of the tangent slope: $m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-\frac{5}{6}} = \frac{6}{5}$

6. Write the equation of the normal line: Again, using $y - y_1 = m(x - x_1)$, substitute $m = \frac{6}{5}$, $x_1 = -1$, and $y_1 = 3$: $y - 3 = \frac{6}{5}(x + 1)$ Simplify: $y = \frac{6}{5}x + \frac{6}{5} + 3$ $y = \frac{6}{5}x + \frac{21}{5}$ So, the equation of the normal line is: $y = \frac{6}{5}x + \frac{21}{5}$

Final Answers:

• Tangent line: $y = -\frac{5}{6}x + \frac{13}{6}$
• Normal line: $y = \frac{6}{5}x + \frac{21}{5}$

Would you like me to verify these or provide a graph of the curve and the lines?

Related Questions:

1. How do we find the slope of a curve geometrically?
2. What is the general method to differentiate an implicit equation like this one?
3. How do tangent and normal lines relate to curvature at a point on a curve?
4. Can we rewrite these equations in standard form instead of slope-intercept form?
5. What is the physical interpretation of a tangent line in an applied problem?

Tip:

When differentiating implicitly, always keep track of $\frac{dy}{dx}$ terms carefully, especially when solving for the slope.