Hey, good to see you again. The problem asks to find the equation of the tangent and normal to the curve $x^2 + y^2 = 11$ at the point $(3, 1)$. Step 1: Find the derivative $\frac{dy}{dx}$ using implicit differentiation. The equation of the curve is $x^2 + y^2 = 11$. Differentiate both sides with respect to $x$: $$ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(11) $$ $$ 2x + 2y \frac{dy}{dx} = 0 $$ $$ 2y \frac{dy}{dx} = -2x $$ $$ \frac{dy}{dx} = -\frac{2x}{2y} $$ $$ \frac{dy}{dx} = -\frac{x}{y} $$ Step 2: Calculate the slope of the tangent at the given point $(3, 1)$. Substitute $x=3$ and $y=1$ into the derivative: $$ m_{\text{tangent}} = \frac{dy}{dx} \Big|_{(3,1)} = -\frac{3}{1} = -3 $$ Step 3: Find the equation of the tangent line. The equation of a line is $y - y_1 = m(x - x_1)$. Using the point $(x_1, y_1) = (3, 1)$ and slope $m_{\text{tangent}} = -3$: $$ y - 1 = -3(x - 3) $$ $$ y - 1 = -3x + 9 $$ $$ y = -3x + 10 $$ Rearrange into the standard form $Ax + By + C = 0$: $$ 3x + y - 10 = 0 $$ Step 4: Calculate the slope of the normal line. The normal line is perpendicular to the tangent line. The product of their slopes is $-1$. $$ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} $$ $$ m_{\text{normal}} = -\frac{1}{-3} = \frac{1}{3} $$ Step 5: Find the equation of the normal line. Using the point $(x_1, y_1) = (3, 1)$ and slope $m_{\text{normal}} = \frac{1}{3}$: $$ y - 1 = \frac{1}{3}(x - 3) $$ Multiply by 3 to clear the fraction: $$ 3(y - 1) = x - 3 $$ $$ 3y - 3 = x - 3 $$ $$ 3y = x $$ Rearrange into the standard form $Ax + By + C = 0$: $$ x - 3y = 0 $$ The equation of the tangent line is $\boxed{3x + y - 10 = 0}$. The equation of the normal line is $\boxed{x - 3y = 0}$. Send me the next one 📸